1. ## Tweed Run

I've been looking at pictures of the Tweed Run. Some of you know what I'm talking about. It seems Brits ride more upright than many of us. More upright means higher CG, and they all seem to be having a splendid good time.

A simple way to explain this issue is with the following concepts:
-High CG- greater range of stability (angular amplitude) but a stronger and slower input is required to correct it.
-Low CG- smaller range of stability (angular amplitude) but weaker and faster inputs are required to correct it.

From a mechanical engineer's point of view, this pretty much sums it
I am fine with that definition apart from the concept of stronger or weaker inputs being required in order to correct losses of stability.

Remember that the corrective input mechanism is the moving of the front wheel from one side of the COG to the other. Because a tall bike falls more slowly the bike will lean less in a given time and so less corrective wheel movement is required. A rider of a less tall bike, reacting in the same time will require more movement. The magnitude of correction required also varies with the speed of the bike and how far back the COG is from the front wheel.
File:Bike weaving.gif - Wikipedia, the free encyclopedia

The notion that it is an input force applied by the bike that corrects the lean is misleading as the force that corrects the bike is that of gravity pulling downwards. The front wheel moves sideways whilst the COG travels in a smooth line. There is a small force that moves the front wheel from side to side but this force does not directly re balance the bike.

To summarize:

-High CG- greater range of stability (angular amplitude), with a slower rate of lean that can be countered more slowly. If countered quickly less front wheel movement will be required.
-Low CG- smaller range of stability (angular amplitude), with a faster rate of lean that needs to be countered more quickly. If countered slowly larger front wheel movement will be required.

3. Originally Posted by I'm suba
I've been looking at pictures of the Tweed Run. Some of you know what I'm talking about. It seems Brits ride more upright than many of us. More upright means higher CG, and they all seem to be having a splendid good time.
They do appear to have a good time. Some commentators say that Geoff was well ahead of the modern tweed wearing trend as this 1984 photo shows.

The tweed run is not typical of British cycling as most UK riders ride in a forward position. The main upright cyclists in Europe are the Dutch, Germans, Belgians and Danes. Though there bikes, unlike Clelands, have slack frame angles, longer wheelbases and low bottom brackets.

4. Originally Posted by smilinsteve
Funny, in the video Graham just posted the guy balances a golf club with head up and head down and says its easier with head down, (COG closer to fulcrum, ie. finger)
Take another look.

He actually says it's easier with the head upwards so that the center of gravity is higher.

5. Originally Posted by I'm suba
I've been looking at pictures of the Tweed Run. Some of you know what I'm talking about. It seems Brits ride more upright than many of us. More upright means higher CG, and they all seem to be having a splendid good time.
I don't look like I'm having much fun, but I'm concentrating I suppose, those rocks are slippy.

6. Originally Posted by GrahamWallace
Take another look.

He actually says it's easier with the head upwards so that the center of gravity is higher.
I watched it again. He says the club head is more easily balanced when it is near your finger. It is shown at time 4:15 to 4:40 in the video.

Edit: But I think he mis-spoke. And looking at the video, he seems to be having more trouble with the club head down.

7. Originally Posted by GeoffApps
I don't look like I'm having much fun, but I'm concentrating I suppose, those rocks are slippy.
Look like a BAMF to me. I really want to try out one of those bikes!

8. Originally Posted by smilinsteve
I watched it again. He says the club head is more easily balanced when it is near your finger. It is shown at time 4:15 to 4:40 in the video.

Edit: But I think he mis-spoke. And looking at the video, he seems to be having more trouble with the club head down.
You are right.

He does in fact say "it seems to me that the club head is more easily balanced with the club head, near!"

I had the sound turned down and thought he said "club head, here!

I eventually find a video that explains the principle and but the demonstrator gets a key word wrong?
But you can see from the other demonstrations that a mass further away from the center of rotation moves more slowly. i.e. it is harder to move.

9. Here's a web page that explains it:

Exhibit Cross Reference - Balancing Stick

10. Bamf?

11. Well done Steve!

Perseverance is the key to understanding counter-intuitive concepts like this.

Bit by bit they start to make sense!

Here's another one to ponder?
When riding up a steep slope you need to keep the weight over the rear wheel to maintain traction. But you also need to keep your weight over the front wheel in order to hold it down.

But how can the weight be over both the front and back wheels at the same time?
Theoretically it can if the weight is an infinite height above the bike.

By the same logic a taller bike should be able to climb a steeper hill than a short one

This also explains why bikes with shorter wheel-bases climb best.

12. Originally Posted by GeoffApps
Bamf?

13. I remember a scene from the seminal film 'On Any Sunday' about the hillclimb called 'The Widowmaker'.
Intuition told those attempting it, year after year, that a huge engine in a mega long wheelbase motorbike with a low low saddle was the way to get to the top.
If I remember rightly, few, if any, succeeded.
Then a chap on a small-engined short wheelbase trials bike, standing on the pegs, slowly tiggled his way up to the top, turned round and came back down.

14. Originally Posted by GrahamWallace
Cor, is that a compliment?

15. Originally Posted by GeoffApps
Cor, is that a compliment?
The meaning of BAMF

Yea, it's a compliment here in the colonies.

Is this the hill climb you are talking about: Hill climb ride up (from movie - On Any Sunday) - YouTube

16. Yes, that's the one.
Clearly, my memory of the film is wrong.
Mind you, the last time I saw the film was in 1970.
Thanks for that link ~ I must get the DVD, it was a most influential film for me.

17. BAMF - yes big compliment.

Geoff and Graham,

Your attitude and responses to the posts in this thread display qualities of intelligent thinking by enthusiastic riders/innovators. Keep up the good work!

As a chiropractor I find Geoff's take on spinal health to be on the money!

18. Originally Posted by BacDoc

Your attitude and responses to the posts in this thread display qualities of intelligent thinking by enthusiastic riders/innovators. Keep up the good work!
It's the English thing. They're farther along on the evolutionary chain.

19. Originally Posted by BacDoc
BAMF - yes big compliment.

Geoff and Graham,

Your attitude and responses to the posts in this thread display qualities of intelligent thinking by enthusiastic riders/innovators. Keep up the good work!

As a chiropractor I find Geoff's take on spinal health to be on the money!
Thanks for that.
I was talking to one of the chiropractors in a group practice which had a stand set up at a cycle show.
He agreed with me that the common 'stretched' posture on a bicycle is bad bad news for most of the body, particularly the spine.
However, if he were to say anyting along these lines to a conventional cyclist he had discovered through experience that, because he was not also a cyclst, he would alienate his clients; not good for them, not good for his income.

I assume you've looked at this

20. Originally Posted by GrahamWallace
But how can the weight be over both the front and back wheels at the same time?
Theoretically it can if the weight is an infinite height above the bike.
Apparently I need educating on this one, as I do not believe it is correct. Could you explain your logic?
Weight distribution is independent of COG height.

[QUOTE=GrahamWallace;9725280By the same logic a taller bike should be able to climb a steeper hill than a short one[/QUOTE]
As the height of the COG increases, so does the rider's need to move his position relative to the axles to account for it. Thus, height will at some point limit the grade that can be climbed without the bike tipping.

21. Originally Posted by meltingfeather
Apparently I need educating on this one, as I do not believe it is correct. Could you explain your logic?
Weight distribution is independent of COG height.
In theory, you can draw a line from the COG to the front or the rear axle, and as you raise the COG that line becomes more vertical. At infinitity, the line is vertical.
But if you want to distribute weight evenly between front and rear, there are easier ways than putting your seat up in the sky. You just center your weight.

Climbing steep hills does require a careful balance between front and rear weight distribution, but 50/50 is not necessarily the best distribution. I think longer top tubes allow for more rider adjustment to move weight back for traction, or to the front for stability, as needed.

As the height of the COG increases, so does the rider's need to move his position relative to the axles to account for it. Thus, height will at some point limit the grade that can be climbed without the bike tipping.
Good point. As the grade gets steeper, the COG moves backwards to a greater degree the higher it is.

22. Originally Posted by smilinsteve
In theory, you can draw a line from the COG to the front or the rear axle, and as you raise the COG that line becomes more vertical. At infinitity, the line is vertical.
Parallel lines do no get closer together (what would be required for even weight distribution) no matter how long you make them. Also, for weight distribution, the horizontal distance from COG to axle is relevant. Height of the COG is irrelevant, unless I'm missing something.

23. Originally Posted by meltingfeather
Parallel lines do no get closer together (what would be required for even weight distribution) no matter how long you make them. Also, for weight distribution, the horizontal distance from COG to axle is relevant. Height of the COG is irrelevant, unless I'm missing something.
Not parallel. Picture you draw a line from the COG to the rear axle, and from the COG to the front axle. So you have a triangle, with a line between the axles as the base. As you raise the COG (at the point of the triangle) the angle at the peak of the triangle gets smaller and smaller...

You never get to parallel, but theoretically you can get that angle as close to zero as you want.

Not that I think this exercise has any practical use. The difference in COG height is too minor in real bike design to be a player in the weight distribution problem. Like you said, its horizontal position that determines it in real life.

24. Originally Posted by meltingfeather
Apparently I need educating on this one, as I do not believe it is correct. Could you explain your logic?
Weight distribution is independent of COG height.

As the height of the COG increases, so does the rider's need to move his position relative to the axles to account for it. Thus, height will at some point limit the grade that can be climbed without the bike tipping.

To repeat my original assertion:

When riding up a steep slope you need to keep the weight over the rear wheel to maintain traction. But you also need to keep your weight over the front wheel in order to hold it down.

But how can the weight be over both the front and back wheels at the same time?
Theoretically it can if the weight is an infinite height above the bike.

By the same logic a taller bike should be able to climb a steeper hill than a short one"

This also explains why bikes with shorter wheel-bases climb best.

Here is my explanation:

The force that causes the front wheel to lift is an equal and opposite reaction to the torque being applied by the rotation of the rear wheel.

In this instance a bike can be considered to be a third class lever with its pivot at the center of the rear axle, the input effort is the torque reaction from the rotation of the rear wheel as applied at the outside of the axle, and the load being the riders weight levered is at the COG. Bear in mind that with a third class lever the input force is always larger than the force applied to its load. Therefore, the further away the COG from the rear axle the smaller the reaction force applied to the COG. And at an infinite distance the force applied at the COG would be zero. In order for the bicycle not to tip forwards or backwards the COG needs to remain within the wheelbase so the only place it can be located at infinity is above the bike.

Of course in reality you can't locate the COG at infinity. Even if you could its inertia would cause the bike to tip backwards as the bike went forwards. However moving the COG as far away from the rear axle whilst keeping it within the wheel-base will reduce the magnitude of the torque reaction. The best place to put the COM is high above the front wheel. Remember that the front wheel is further up the hill than the rear wheel so the wheel-base will have shortened relative to the horizontal. Due to the inertia of the COG any acceleration force applied to the will increase rear wheel traction but can also cause the bike to tip backwards. So smooth pedaling is essential. This can be aided by the use of elliptical gears that also allow the cadence to be lowered whilst reducing the chances of rear wheel slip.

Its never fails to amaze me just how complicated is the physics of the simple machine that we call the bicycle.

To help with the key concepts here are two videos:

In this one of unicyclists cycling uphill you can see how unicyclists have to lean forvard as they pedal in order to counter the combined effect of the torque reaction and the inertia of their COG.

And towards the end of this this one, the Third Class Lever is explained.

25. Originally Posted by smilinsteve
Climbing steep hills does require a careful balance between front and rear weight distribution, but 50/50 is not necessarily the best distribution. I think longer top tubes allow for more rider adjustment to move weight back for traction, or to the front for stability, as needed.
With a longer top tube your upper body is stretched in order to reach the handlebar.
Your upper body is what you use to transfer weight fore and aft.
A stretched upper body has only limited capacity for moving forwards and backwards.
Therefore, a shorter toptube and reach does allow the flexibility.

26. So, drum brakes? What kind of drive train are you using?

27. Originally Posted by smilinsteve
Not parallel. Picture you draw a line from the COG to the rear axle, and from the COG to the front axle. So you have a triangle, with a line between the axles as the base. As you raise the COG (at the point of the triangle) the angle at the peak of the triangle gets smaller and smaller...

You never get to parallel, but theoretically you can get that angle as close to zero as you want.

Not that I think this exercise has any practical use. The difference in COG height is too minor in real bike design to be a player in the weight distribution problem. Like you said, its horizontal position that determines it in real life.
The horizontal positions available is limited as you cannot move the COG outside of the wheelbase. Therefore the only way you way you can reduce the leverage effects of rear wheel torque reaction is by moving the COG upwards.

The practical effects of this are indeed small. However, if you wanted to win a hill climbing competition at best it could increase the incline your bike could climb by the same proportion as the reduction in mechanical advantage caused by increasing the length of the 3rd angle lever. The distance between the rear axle and the COG.

44% hill climb.

28. Originally Posted by leeboh
So, drum brakes? What kind of drive train are you using?
Back in the early 1980s, Cleland bicycles were fitted with Leleu 'floating cam' drum brakes. Apart from those made by Dave Wrath-Sharman, as far as we are aware, the Leleu brakes were the only floating cam drum brakes on the world market. All other drum brakes were, and are, fixed cam. This makes them unsuitable for extreme weather conditions and off-road use.
The Leleu company in Lyons, France, closed down several years ago.
Modern Cleland style bicycles use roller brakes. These are also in the hub and function completely differently to drum brakes. Roller brakes provide better modulation than disc brakes, they are cheap as chips, require virtually no maintenance (the pair I currently use are six years old and have had no more than 12 minutes maintenance in that period, and required no replacement parts). Moreover, they require grease to function and are thus completely impervious to water, so river crossings present no issues at all. The downside is that they are relatively heavy; with all those plus points, I'm quite happy to carry a few extra pounds.

Pictorial details of the drivetain can be seen earlier in this thread. Further information can be found on the Cleland website. The drivetrain is fairly unusual, so as a general topic it is very broad. Once you have done your research, should you have any questions about specific areas, I'd be happy to respond.

29. Originally Posted by GeoffApps
Back in the early 1980s, Cleland bicycles were fitted with Leleu 'floating cam' drum brakes. Apart from those made by Dave Wrath-Sharman, as far as we are aware, the Leleu brakes were the only floating cam drum brakes on the world market. All other drum brakes were, and are, fixed cam. This makes them unsuitable for extreme weather conditions and off-road use....
I've used drum brakes and roller brakes offroad. A properly greased Shimano roller brake is much better than most people realise. I'd agree about the comments on the advantages of floating cam drum brakes, but Sturmey-Archer are now producing a 90mm drum which has more than adequate power (still not as powerful as a good disk though). However we're talking brakes that function longterm in foul conditions, and my experience is that disks don't do that without consuming expensive pads and needing maintenance.

I've done 24 hour races in foul conditions with S-A drum brakes. I used them not because they are great brakes but because they were adequate brakes. Their great advantage is that in conditions where some competitors were changing their pads several times (as I had in earlier races) I was losing no time in the pits. When it's pitch black and the temp is around -10ºC I can take over 15 minutes an end to do what is usually a quick job. I reckon that was worth 1 lap overall, better than I could pick up by being fitter.

This year's race was done on the larger S-A 90mm drums, which were much nicer to use because they didn't need the death grip to work on fast downhills. When stripped down after the race, there was negligible wear on the brake shoes and the internals were clean.

This was one section before the race. It turned into mud soup and a brake grinder within the first lap.
[IMG]

Like this

After the race. (The mudguards were to reduce the chance of hypothermia from being soaked more than to keep the mud off)
[IMG]

[IMG]

Internal pics
[IMG]

Oh, btw, before I get pilloried for my unfashionable red bar ends in the wrong position, I'd better point out that they are so if I have to turn my bike upside down for trail repairs they will keep the bar off the ground and thus I don't have to remove my lights etc from the bars.

30. Originally Posted by GrahamWallace

To repeat my original assertion:

When riding up a steep slope you need to keep the weight over the rear wheel to maintain traction. But you also need to keep your weight over the front wheel in order to hold it down.

But how can the weight be over both the front and back wheels at the same time?
Theoretically it can if the weight is an infinite height above the bike.

By the same logic a taller bike should be able to climb a steeper hill than a short one"

This also explains why bikes with shorter wheel-bases climb best.

Here is my explanation:

The force that causes the front wheel to lift is an equal and opposite reaction to the torque being applied by the rotation of the rear wheel.

In this instance a bike can be considered to be a third class lever with its pivot at the center of the rear axle, the input effort is the torque reaction from the rotation of the rear wheel as applied at the outside of the axle, and the load being the riders weight levered is at the COG.
Bear in mind that with a third class lever the input force is always larger than the force applied to its load. Therefore, the further away the COG from the rear axle the smaller the reaction force applied to the COG. And at an infinite distance the force applied at the COG would be zero. In order for the bicycle not to tip forwards or backwards the COG needs to remain within the wheelbase so the only place it can be located at infinity is above the bike.

Of course in reality you can't locate the COG at infinity. Even if you could its inertia would cause the bike to tip backwards as the bike went forwards. However moving the COG as far away from the rear axle whilst keeping it within the wheel-base will reduce the magnitude of the torque reaction. The best place to put the COM is high above the front wheel. Remember that the front wheel is further up the hill than the rear wheel so the wheel-base will have shortened relative to the horizontal. Due to the inertia of the COG any acceleration force applied to the will increase rear wheel traction but can also cause the bike to tip backwards. So smooth pedaling is essential. This can be aided by the use of elliptical gears that also allow the cadence to be lowered whilst reducing the chances of rear wheel slip.
I am not seeing this at all.

Third class lever? that would mean the force is in between the load and fulcrum. If the load is the COG and the fulcrum is the axle, what is the force and where is it applied? You say:

the input effort is the torque reaction from the rotation of the rear wheel as applied at the outside of the axle
Well, the axle is attached to the bike by freely rotating ball bearings so there is no torque transfered from the wheel to the axle or the axle to the frame.

The ground force creates a force reaction, not a torque, at the rear axle pushing the bike forward.

So the force is at the fulcrum, and if the force is at the fulcrum it creates no rotation around the fulcrum.

So what makes the front wheel lift?

In a climb, like in that 44% grade picture, the only contact points of the rider to the bike are at the handlebar and at the pedals. If the bike and rider are at equilibrium with both tires on the ground, then you can raise that COG to infinity (along the force vector) without effecting that equilibrium.

Now, suddenly the front wheel comes up, so one of these things must have happened.
1. An upward force from the ground (bump).
2. A rearward weightshift such as caused by acceleration from a pedal stroke.
3. Pulling up on the handlebars.

That's all I can think of. And height of the COG doesn't effect any of them unless changing the height also changes the for/aft weight distribution.

31. Originally Posted by meltingfeather

As the height of the COG increases, so does the rider's need to move his position relative to the axles to account for it. Thus, height will at some point limit the grade that can be climbed without the bike tipping.
Your above statement is true if the COG is being accelerated by drive forces emanating from the rear wheel. It would not be true if the COG is moving at a constant velocity.

Of course with a low frequency pulsed drive of a bicycle some acceleration and deceleration of the COG is inevitable. So keeping the drive smooth as possible is essential. Hence the use of elliptical gear rings. It is only the reaction forces to any acceleration that can cause the front wheel to lift. The question is where to best position the COG in order that its weight can best counter these reaction forces.

Remember that to move it forwards will reduce the traction of the rear wheel.

And to move it backwards will of itself, will reduce the leverage that the COG has over the reaction forces.

To move the COG upwards along the diagonal axis between the rear axle and the COG will increase the leverage that the COG has over the rear wheel torque reaction. Whilst the reaction force due to COG inertia, remains unchanged.

32. Originally Posted by GrahamWallace
...Of course with a low frequency pulsed drive of a bicycle some acceleration and deceleration of the COG is inevitable. So keeping the drive smooth as possible is essential. Hence the use of elliptical gear rings. It is only the reaction forces to any acceleration that can cause the front wheel to lift. The question is where to best position the COG in order that its weight can best counter these reaction forces...
I hadn't thought of that advantage of using elliptical rings. On really rocky steep hills, I've usually found the inability to keep the front wheel down stops me more often than the loss of puff or traction. I'll have to give it a try and see if it works for me.

Edit: just been to HIghpath site, not currently available. Anywhere you recommend for these?

33. Originally Posted by GrahamWallace
Your above statement is true if the COG is being accelerated by drive forces emanating from the rear wheel. It would not be true if the COG is moving at a constant velocity.
It is true even if moving at constant velocity. The comment is specifically in regard to a bike on grade. At some point the bike will tip due to the grade moving the COG rearward of the rear axle, even if standing still. Increasing the grade as well as raising the COG make tipping due to acceleration easier.
Originally Posted by GrahamWallace
The question is where to best position the COG in order that its weight can best counter these reaction forces.

Remember that to move it forwards will reduce the traction of the rear wheel.

And to move it backwards will of itself, will reduce the leverage that the COG has over the reaction forces.

To move the COG upwards along the diagonal axis between the rear axle and the COG will increase the leverage that the COG has over the rear wheel torque reaction. Whilst the reaction force due to COG inertia, remains unchanged.
Well put, and exactly my point. Without the forward movement, raising the COG is counter productive to weight distribution on a bike on a grade.

34. Originally Posted by NEPMTBA
And why not Horses?

They graze as they go, water at any stream, can carry more gear, and are faster...LOL
An Army commander was very influential for the start-up of bicycle mounted soldiers, had lots of pull! Not only that, bicycle mounted troops were popular all over Europe and Africa for many years. The Swiss were very much involved and still are to a degree.

poikaa

35. Originally Posted by meltingfeather
Parallel lines do no get closer together (what would be required for even weight distribution) no matter how long you make them. Also, for weight distribution, the horizontal distance from COG to axle is relevant. Height of the COG is irrelevant, unless I'm missing something.
I have rethought what I wrote earlier, and agree that the height of the COG is irrelevant to the weight distribution.
And I would appreciate it, Melting Feather, if you would slap my silly as\$ back into place when I say stupid crap

The COG can be represented by a point and it creates a force vector directed from that point to the center of the earth. That vector is on a line that goes up into space and no matter where you move the COG up that line, the force vector to the ground does not change, so the weight distribution doesn't change.

36. I have a really short top-tube, short wheelbase bike with an oval chainring. I agree with Geoff and Graham re climbing balance. Also on the trails-style handling benefit of a higher COG, a taller front end and more upright position puts me higher on teh bike when stood up and it has real benefits on slow, technical terrain comaperd to conventional bike design. Personally I prefer the way a low-BB bike can corner at speed and the planted feel it gives on a steep descent, but bikes are all about trade-offs and compromises - often the bikes that don't need to be too compromised are the best for some riders. Depends on how / what you ride.

BTW, I've ridden that slope pictured earlier as I live + ride in the same area. It's steep, but not at-the-limit crazy-steep imo. It's long, uneven and traction is low though, so as a test of control it's a good one!

Always interesting to read about these bikes, so thanks to Geoff and Graham for sharing your take on bike design. I think it's only the fact that common MTB culture has taken the direction it has that has made this style of bike marginal in its appeal, since the theory and 'riding ideal' that it's based on is excellent.

37. Maybe see you at the Birthday Ride this December?

38. Here is the problem in diagrammatic form.

The vertical broken line represents the center of the wheelbase.

The red arrow represents the rear wheel torque but is not shown to scale. (If shown to scale is would be more than twice the size of longest reaction force shown).

I have used black dots to represent three differently placed COGs, Attached an arrow to each black dot that shows the direction and relative magnitude of the torque reaction acting at each COG. (The top COG though not achievable on a standard bicycle is nonetheless, theoretically interesting).

Each dotted line connecting the rear wheel to a COG represents the central axis of a virtual third class lever that transmits the anti clockwise torque reaction forces. In reality these forces would be transmitted through the bike and the body of the rider. (Remember that with a third class lever, the longer the lever, the less force is transmitted to the COG).

Not shown is:
* the large gravitational force acting vertically downwards on each COG.
* the drive forces that act upwards and forwards from the rear axle to each COG.( These forces act as an equal and opposite reaction to the rear wheel torque They push the COG both onwards and upwards).
* the reaction forces that would be generated by sudden accelerations of the COG caused by pedaling that is not smooth, (these will act at the GOG in equal and opposite direction and magnitude to any acceleration or deceleration).

But what COG position will allow you to climb the steepest hill?

If your not confused by now you're not trying hard enough

Missing info: is the mass of the bike and rider and the coefficient of friction of the rear tire.

39. Maybe see you at the Birthday Ride this December?
The Wendover ride? Would love to, thanks.

edit to add, Graham doesn't do a lecture in the pub after does he..? Just kidding. Keep at it and I'll keep reading.

40. That's a nice drawing Graham.
I still don't think we are on the same page.
I do appreciate discussing it with you.

What I'm thinking.

- The red line is not a torque, it is the force vector caused by the tire pushing on the ground, and the equal and opposite vector is acting at the rear axle pushing the bike forward.

- There is no "torque reaction" at each COG. The only forces on this system are red arrow, the forward force reaction at the rear axle, the downward force from the weight of the bike and rider, and the normal force up from the ground on the tires that is countering the weight.

- When the bike accelerates from the red arrow force, if the bike and rider act as 1 rigid body, the position of the COG doesn't effect the motion. The only thing that matters is that the COG is in a place to provide sufficient force on the tires for traction in the rear and for stability of the front.

-Normally, the bike and rider do not act as one rigid body. Acceleration will cause the bike to squirt out from under the rider, causing a relative weight shift backwards (the inertia of the rider is greater than that of the bike). A higher COG potentially makes this worse because the there is a longer lever arm from the COG to where the force is applied to the rider (his feet and hands if he is standing on the pedals).

- In your picture, the answer to your question "which COG position will allow you to climb the steepest hill?" the answer is the most forward one if front wheel lift is the limiting factor, but in general it is the one that gives you the right balance of rear traction and front stability. This is detemined by the for-aft position of the COG only. I don't see how the height makes any difference at all.

- I still see nothing analagous to a class 3 lever, since I disagree that there is a force (or "torque reaction") at the COG that you have drawn.

41. Originally Posted by GrahamWallace
Remember that with a third class lever, the longer the lever, the less force is transmitted to the COG.
This argument has a converse that equally and oppositely counters it.
As you move the COG away from the axle you also decrease the force/inertia effect required to impart torque/rotation about the axle by increasing the lever arm (the same mechanism by which the torque reaction is being reduced).
In your own example of COG at infinity, when the "force transmitted to the COG," as you like to put it, is zero, the bike tips upon first input, which is not an equalization of weight distribution as you claimed, but the opposite extreme. The basic 3rd class lever example falls short in that, on a bicycle, the pivot point is free to move, and so, with a COG at infinity, no motion or force input at the COG is required for the relative rotation of the COG about the axle, which occurs due to movement of the pivot point, and the net effect is movement of the COG behind the rear axle (and tipping).

Heightening the COG exacerbates the weight distribution effects of acceleration and grade.
As you move the COG forward you balance and counter the effects of both forward acceleration and grade.

I think that the raising of COG, which you are making as an argument for effective hill climbing, is nothing but a side-effect of moving forward, which on a bicycle requires standing.
Originally Posted by GrahamWallace
If your not confused by now you're not trying hard enough
Interesting dialogue, to be sure.
You Brits certainly are masters of adverse conditions riding. The terrain around here gets destroyed if ridden in those conditions and if I posted a photo of riding local terrain in that condition I'd get tarred and feathered by any local mtb'ers that came across it.
Not to mention our climate can be described as long periods of drought punctuated by brief interludes of flash flooding.

42. Originally Posted by smilinsteve
That's a nice drawing Graham.
I still don't think we are on the same page.
I do appreciate discussing it with you.
We may not be "on the same page" but are both seekers of the same truth.

Steve you make some valid and interesting points there. However, before I reply in detail I would like to clarify whether you believe that torque reaction forces are involved at all?

Here's a video of a Segway driving up and down a steep slope. Note the Segway is moving at a constant velocity and is not accelerating. However the rider can be seen to lean forwards markedly. Why does she need to do this and why does she not fall forward. Does it make sense that a faster rider or a shorter rider of the same weight would need to lean forwards even more?

There is clearly some leverage involved in stopping the rider from falling forwards.. Could this leverage be categorized as the class three? And could the leverage result from the torque reaction to the motor/gearbox?

Why doesn't the rider need to lean when going downwards?

Ideally we need some good footage of a unicyclist riding up a steep slope, but so far, I haven't found a good example.

43. Originally Posted by meltingfeather
This argument has a converse that equally and oppositely counters it.
The end game here is to arrive at some kind of consensus that we can agree on.
There is no value in me having a theory if it won't stand up to scrutiny.

We are treading new ground here, I don't know of any discussion of this topic in the "bicycle science" textbooks or Wikipedia.
Take a look at the Segway clip that I have posted, what do you think?.

My I ideas are based on testing various bike designs on my local 44% hill and then coming up with a rational of my own for what I have found. I have also built radio controlled models to test out ideas. I am hoping that what I learn from this conversation may feed into future bike designs.

I will carefully consider your and Steve's thoughts and then report back in detail.

44. I'm reminded of a purported quote from a former Australian Prime Minister:

"That may work in practise, but it doesn't fit my theory" or words to that effect.

The theories here all make sense in themselves, but the QED here is that Graham is climbing these hills when most can't. Finding out whether that is because he has monster legs and an incredibly smooth technique, or because of the bike design is the fun part.

45. Originally Posted by GrahamWallace
We may not be "on the same page" but are both seekers of the same truth.

Steve you make some valid and interesting points there. However, before I reply in detail I would like to clarify whether you believe that torque reaction forces are involved at all?

Here's a video of a Segway driving up and down a steep slope. Note the Segway is moving at a constant velocity and is not accelerating. However the rider can be seen to lean forwards markedly. Why does she need to do this and why does she not fall forward. Does it make sense that a faster rider or a shorter rider of the same weight would need to lean forwards even more?

There is clearly some leverage involved in stopping the rider from falling forwards.. Could this leverage be categorized as the class three? And could the leverage result from the torque reaction to the motor/gearbox?

Why doesn't the rider need to lean when going downwards?

Ideally we need some good footage of a unicyclist riding up a steep slope, but so far, I haven't found a good example.

Graham you are really making me do my homework! I don't know anything about Segways, but now you give me something else to figure out.
That's ok. I've been involved in these physics discussions in this forum for a couple of years (and have bumped into Melting Feather in a few of them), and I think I have learned more about physics at MTBR then I ever did in college (or close anyway).
I will get back to you on this. I hope however, that I have enough fun, and a good ride, this weekend, that will keep me from spending too much time on the computer.

46. Originally Posted by GrahamWallace
Here is the problem in diagrammatic form.

> SNIP <

But what COG position will allow you to climb the steepest hill?

If your not confused by now you're not trying hard enough

> When the tyres loose contact! When this happens you will also be in a controlled or uncontrolled crash! <

Missing info: is the mass of the bike and rider and the coefficient of friction of the rear tire.
Some sort of "zero-sum-game" one plays with gravity.... the bike becomes a tool to win and the better you know your "tool", the greater your "edge"!

poikaa

47. Originally Posted by smilinsteve
Graham you are really making me do my homework! I don't know anything about Segways, but now you give me something else to figure out.
Here's an explanation of Segway technology from Wikipedia:

"The dynamics of the Segway PT are similar to a classic control problem, the inverted pendulum. The Segway PT (PT is an initialism for personal transporter while the old suffix HT was an initialism for human transporter) has electric motors powered by Valence Technology phosphate-based lithium-ion batteries which can be charged from household current. It balances with the help of dual computers running proprietary software, two tilt sensors, and five gyroscopic sensors developed by BAE Systems' Advanced Technology Centre[5]. The servo drive motors rotate the wheels forwards or backwards as needed for balance or propulsion. The rider controls forward and backward movement by leaning the Segway relative to the combined center of mass of the rider and Segway, by holding the control bar closer to or farther from their body. The Segway detects the change in the balance point, and adjusts the speed at which it is balancing the rider accordingly. On older models, steering is controlled by a twist grip on the left handlebar, which simply varies the speeds between the two motors, rotating the Segway PT (a decrease in the speed of the left wheel would turn the Segway PT to the left). Newer models enable the use of tilting the handle bar to steer"
Segway PT - Wikipedia, the free encyclopedia

The best example a "torque reaction"/third class lever machine, that I know of, is the classic dragster design. The long front end being one big reaction arm to the rear wheel torque..
.
The COG of the dragsters engine is directly ahead of the rear axle so as to eliminate the chance of the engines inertia lifting up the nose and then flipping the dragster backwards. But in order to gain the maximum rear wheel traction the COG of the dragster needs to be as close to the rear axle as possible. The problem is that the torque reaction is powerful enough on its own, to lift the engine and flip the dragster backwards. To solve this problem by adding a large extra forward weight, would only serve to slow down the dragster. And moving the engine further forward, away from the axle, would reduce rear wheel traction.

So instead they add a long and lightweight third class lever reaction arm onto the front. By placing a small weight at the nose end of this arm will then create enough down-force to counter the torque reaction. This works in the same way that a weight in an outstretched arm feels heavier than the same weight when your arm is bent.

One thing to note regarding the video of the hill climbing Segway is that no bicycle under its own power, could ever ride up a slope that steep. The rear wheel torque of a bicycle would be nowhere near smooth enough. To make big inroads into the maximum incline that can be ridden on a bicycle, the problem of the pulsed nature of rear wheel torque would need to be addressed.

48. Originally Posted by Poikaa
Some sort of "zero-sum-game" one plays with gravity.... the bike becomes a tool to win and the better you know your "tool", the greater your "edge"!

poikaa
For me, riding up challenging hills is inexplicably compulsive.
Though I have never been interested in racing on mountain bikes I love to take on and hopefully defeat hills. It's part of my psyche, Like riding in bad weather. I seem to be more interested in defeating nature than other people. It's a form of insanity' It's very hard work and these hills would be easier and quicker to conquer on foot. And most of my climbs are attempted when no one is watching to see me either succeed or fail. I guess it's good exorcise and less dangerous than most other things that people do on mountain bikes.

I do however recommend that people who try this practice ways of getting off quickly when things go wrong. As not knowing how to react when the rear wheel suddenly spins out on a 44% slope could result in you returning down the hill head first, with the bike landing on top of you. I am now equally as good at bailing out, as I am as getting into trouble in the first place.

49. Originally Posted by Velobike
"That may work in practise, but it doesn't fit my theory" or words to that effect.

The theories here all make sense in themselves, but the QED here is that Graham is climbing these hills when most can't. Finding out whether that is because he has monster legs and an incredibly smooth technique, or because of the bike design is the fun part.
I know which of my bikes can climb my 44% test slope and I know those that can't. I even keep information of the relative success rates of the good climbers. The problem is that all my 5 bikes are different.In order to identify exactly what made then climb better I would have to change one parameter at a time and then test the slightly altered bikes in identical conditions. I would also need to repeat the rides many times and record and average out the runs to take account of variations in the exact path taken.

Though I am mad enough to do all this, in reality I neither have the time or the money.

In mountain biking like many other sports the main driver for innovation is racing. The fact that we don't have a category of 'Uphill Racing' means that very little effort has gone into improving the climbing characteristics of mountain bikes.

50. Originally Posted by GrahamWallace
The end game here is to arrive at some kind of consensus that we can agree on.
There is no value in me having a theory if it won't stand up to scrutiny.

We are treading new ground here, I don't know of any discussion of this topic in the "bicycle science" textbooks or Wikipedia.
Take a look at the Segway clip that I have posted, what do you think?.

My I ideas are based on testing various bike designs on my local 44% hill and then coming up with a rational of my own for what I have found. I have also built radio controlled models to test out ideas. I am hoping that what I learn from this conversation may feed into future bike designs.

I will carefully consider your and Steve's thoughts and then report back in detail.
This is an interesting discussion, and certainly something I need to think more about; I will.
The Segway context is something i need to digest a bit.
The dragster example has a couple of key differences: weight as a critical design driver being primary.

51. Here's a video of a Segway driving up and down a steep slope. Note the Segway is moving at a constant velocity and is not accelerating. However the rider can be seen to lean forwards markedly. Why does she need to do this and why does she not fall forward. Does it make sense that a faster rider or a shorter rider of the same weight would need to lean forwards even more?

There is clearly some leverage involved in stopping the rider from falling forwards.. Could this leverage be categorized as the class three? And could the leverage result from the torque reaction to the motor/gearbox?
I'm not so good at the numbers side of all this, I seem to 'see' more complex mechanics problems in my mind or feel it on the bike but struggle to work through the actual maths in detail. So I can't put the sums together here without a headache and a few hours, if at all, but I think yes, surely you lean forwards due to the torque reaction? Like countering a rearward thrust you feel under acceleration. Maybe I'm simplifying this due to not fully grasping the problem though.
When climbing over 25-30% in a low gear, despite the force on the crank and your fwd COG, your gearing must be so low that you can wheelie out easily if you move weight back a little. The force into the crank is bodyweight times 'x' (x is your strength adding to your weight) but the gearing multiplies the torque more than your strength or body position range on the bike multiplies your weight? (--edit to clarify - I mean when it gets limitingly steep. At that point you either can't pedal as you're trying so hard to balance your weight, or can't prevent loop-out if you can pedal hard enough). Is that not a torque reaction in the same way as the segway, the segway just simplifies it?

All I do know is that when I've 'sessioned' steep climbs with friends, what stops us is the front wheel lifting over a bump and upseting balance, looping out, or stalling against uneven ground.. or fitness eventually / rapidly! My weight is at that point, it'll always be somewhere between the axles depending on what the wheels are going over and technique and skill count for all (ie weight in right place at right time) and the variance in bikes much less. Having said that, I'm talking about fairly conventional bikes, from std 26",29" to a Jones at the most extreme example. There's plenty of times when a really high BB, short cranks and short wheelbase would help keep the power going, but that same set up wouldn't be so good in other situations.

I guess what I'm thinking is that good technique is always needed and a bike can't do all things well. It has to be biased well toward a riding style, but always needs a lot of dynamic input - that input is the beauty of a bicycle as well as what complicates these physics questions! You can make it easier for that input to effect the bike but much depends on the rider.

Velobike - try Goldtec in the Uk, maybe via Tazzy on STW. Oval rings at a lower cost than Rotor. Not the most durable ime so far, mainly due to uneven pressure on the teeth and burring at the larger raduis area, but not bad at all, I like them. Great on the SS.

52. Originally Posted by meltingfeather
The dragster example has a couple of key differences: weight as a critical design driver being primary.
The dragster example was not intended to illustrate how forces behave in a bicycle. Just an example of how torque reaction and third class levers are used on other machines.

53. Aren't we talking about a couple of inches here? It seems to be a moot point.

54. Originally Posted by wmac
Aren't we talking about a couple of inches here? It seems to be a moot point.
Yes and no!
When it comes to the ideal positioning of the riders Center of Gravity we are talking about only making marginal gains. Which is only in the, impress your friends with your skill territory. But if you compare the steepness of the slope the Segway is climbing, described as 45 degrees (100%) but looks more like 80%. We are in the region of twice the incline that you can ride on a mountain bike. So making a bike that could climb as well as a Segway, would be a big step forwards.

Imagine being able to do this on a mountain bike:
Segway Steep hill with wipeout - YouTube

Notice that the crash at the end happens because the motors don't have enough torque to counter the forward lean.

55. Wow, I am surprised that Cloxxki and David Copperfield have not chimed in on this one.

Your style of riding is far too regionalized. Those bikes would suck in the dez and just about everywhere were you need to make time between features on even terrain.

If you so want recognition for inventing 29" wheels you should nominate Geoff to the MTBOF (mtnbikehalloffame.com), stuff the ballot box with your band of merrymen, and then bring a few bikes to Vegas for the induction next year.

They nominated a cigarette smoking non bike riding guy this year by that method so you stand a chance of getting him in to the club but I would wager no chance on the trails. Not saying you couldn't ride them but you won't be able to hang.

56. I am afraid that the style of voting for the MTBHOF is far too regionalized.

57. Originally Posted by Bigwheel
...Your style of riding is far too regionalized. Those bikes would suck in the dez and just about everywhere were you need to make time between features on even terrain.

If you so want recognition for inventing 29" wheels you should nominate Geoff to the MTBOF (mtnbikehalloffame.com)....
I very much doubt he needs to chase recognition. He already has that where it matters, except maybe in the small region called USA.

As for the invention of big wheels, perhaps you could enlighten us as to who you believe had the first application of them to purpose built all terrain bicycles. (And the big wheeled fat tyred bikes of 100 years ago don't count because the geometry of those bikes was road oriented).

58. Segway video - it shows a guy leaning forward as he goes up the hill at constant speed. My understanding of the segway is that you signal it to move by leaning forward. Also, almost all the weight of the Segway is at the bottom, so the center of mass of the seqway plus rider is very low. He is leaning just barely forward, with center of mass still near the tire contact patch. The forward lean causes the Segway to tilt, but not enough to tip it forward. I don't thing there is any torque reaction involved in this.

Leaning forward to me is similar to leaning forward on a skateboard. Going up hill, hitting a bump or decelleration causes a weight shift back, so you lean forward in anticipation of any deceleration.

Conversely, when going downhill you lean back because a deceleration will cause a forward weight shift.

59. Dragster: I might be missing the whole concept of "torque reaction", but again I don't see it in this example.

I don't think the front end of the dragster is long and weighted to counter a torque reaction. I think that front end lift comes from air lift and bumps in the road. In other words, on a perfectly smooth flat ideal track in a vacuum, the front end won't lift! (?)

(The force of the engine is acting through the rear axle, which is higher than the front axle).

60. Torque reaction - Torque reaction is the rotational equivalent of Newton's equal and opposite force reaction. Some good examples of this are a helicopter, where rotation of the blades tends to cause an opposite rotation of the helicopter body. Or a pneumatic nut driver, that causes a torque reaction that twists the tool in the operators hand.
I don't see any torque reaction in the bike climbing a hill.

61. Originally Posted by james-o
When climbing over 25-30% in a low gear, despite the force on the crank and your fwd COG, your gearing must be so low that you can wheelie out easily if you move weight back a little. The force into the crank is bodyweight times 'x' (x is your strength adding to your weight) but the gearing multiplies the torque more than your strength or body position range on the bike multiplies your weight? (--edit to clarify - I mean when it gets limitingly steep. At that point you either can't pedal as you're trying so hard to balance your weight, or can't prevent loop-out if you can pedal hard enough). Is that not a torque reaction in the same way as the segway, the segway just simplifies it?
I still do not see any torque reaction coming into play in the bicycle hill climb. As the grade gets steeper, center of gravity moves back. The front wheel has less weight on it. That makes it more likely to lift if you pull on the handlebars, or hit a bump with the front tire.

62. Originally Posted by smilinsteve
Segway video - it shows a guy leaning forward as he goes up the hill at constant speed. My understanding of the segway is that you signal it to move by leaning forward. Also, almost all the weight of the Segway is at the bottom, so the center of mass of the seqway plus rider is very low. He is leaning just barely forward, with center of mass still near the tire contact patch. The forward lean causes the Segway to tilt, but not enough to tip it forward. I don't thing there is any torque reaction involved in this.

Leaning forward to me is similar to leaning forward on a skateboard. Going up hill, hitting a bump or decelleration causes a weight shift back, so you lean forward in anticipation of any deceleration.

Conversely, when going downhill you lean back because a deceleration will cause a forward weight shift.
Segway weight 83lbs

Segway weight capacity: 260 pound (118 kg) rider and cargo.

That would put the COG of rider and Segway around the top of the riders legs.

HowStuffWorks "Segway: Car Replacement?"

63. Originally Posted by smilinsteve
Dragster:
I don't think the front end of the dragster is long and weighted to counter a torque reaction. I think that front end lift comes from air lift and bumps in the road. In other words, on a perfectly smooth flat ideal track in a vacuum, the front end won't lift! (?)
Would then not the simple solution be to fit a short front end and suspension?

Originally Posted by smilinsteve
(The force of the engine is acting through the rear axle, which is higher than the front axle).
The key distance is to have the COG at the same height as the rear axle as this would cancel out the tendency of the inertia focused at the COG to lift the front. But even then a mechanism is still required to cancel the torque reaction.

64. Originally Posted by smilinsteve
Torque reaction - Torque reaction is the rotational equivalent of Newton's equal and opposite force reaction. Some good examples of this are a helicopter, where rotation of the blades tends to cause an opposite rotation of the helicopter body. Or a pneumatic nut driver, that causes a torque reaction that twists the tool in the operators hand.
That's correct.

Originally Posted by smilinsteve
I don't see any torque reaction in the bike climbing a hill.
Newton's Third Law: "To every action there is always an equal and opposite reaction: or the forces of two bodies on each other are always equal and are directed in opposite directions".

This also applies to rotational leverage forces known as torque, where a clockwise torque will always produce an "equal and opposite" anti-clockwise reaction.

If the rear wheel has torque, there must be an equal and opposite force transmitted to the rest of the bike. However, the exact magnitude of this force will depend on the distance away from the axle that you measure it.

65. smilinsteve + Graham - is a torque reaction part of how we start a wheelie? Partly deliberate weight shift, partly using a low-ish gear to initiate that weight shift from the torque at the rear wheel. And popping wheelies is one thing that stops us climbing steep hills seated - mix of COG change due to the hill and the gearing producing more torque.
And the dragster has a long wheelbase partly to help it steer straight, I thought. They have wheelie-bars to help against the torque reaction. I'm not sure why they don't have more downforce up front via aerofoils, probably since teh torque is highest when speed is lowest. But I'm a bit lost as to the original point of all this anyway )

If you make a great hill-climber for 40% + grades you probably have a bike that won't ride well on other trails. I saw a bike with a 20" wheel once in an off-road hill-climb event .. kept the COG in the right place as well as good pedalling position - a rubbish all-round MTB though! The bush-whacking do-all in a trials-style ethic of a Cleland is great, but I think it'd need to do so without comproming flow at higher speeds too much to gain popularity. Niche appeal is no bad thing, but marginalisation is a shame.

Velobike, I think those old roadsters aren't that far off the geometry of a Jones in some areas, and closer again to many really early MTBs? Exact numbers are different but the basis is similar, long rake forks, slack angles, long wheelbases and grips close to in-line with the steerer etc. I rode a early 1900s roadster once, very briefly, but it felt pretty good. I think people pointing out that no-one really 'invented' 700c off-road bikes have a valid point, they evolved slowly. To me, the early days of the Tour de France was the starting point of off-road riding - big mountains, dirt and mud roads, solo riders vs the terrain - the TDR isn't much different now. Culturaly it took the Marin Co downhillers to grab popular attention,and also for road riding to be something established that there could be an alternative to. Sometimes things have to settle in one area for some people to see a real difference in something new, or react against it and forge ahead in a different direction. But some rider/designers were pretty influencial between the 'birth of MTB' and now, and would have been more so if things like tyres that need a lot of £ to make hadn't influenced MTB direction. I'm just waffling now, no real point to make, just my thoughts on why these bikes and Geoff's ideas are recognised.

I remember seeing a Highpath MTB in a magazine in the 80s and thinking the go-anywhere ideas in its design were exciting. No less so than the clunkers from Marin Co, just a different appeal.

66. Originally Posted by james-o
And popping wheelies is one thing that stops us climbing steep hills seated - mix of COG change due to the hill and the gearing producing more torque.
Yes, that's why it's pointless to fit a 15 tooth grannyring. The torque produced would either spin out the back wheel or tip you over backwards.

Originally Posted by james-o
I thought. They have wheelie-bars to help against the torque reaction. I'm not sure why they don't have more downforce up front via aerofoils, probably since teh torque is highest when speed is lowest.
Long nose dragsters don't need wheelie-bars, as there is no need to have two mechanisms that do similar things. Modern dragsters do have aerofoils, though unlike torque reaction arms they only create downforce when the dragster is moving fast.

Originally Posted by james-o
But I'm a bit lost as to the original point of all this anyway )
This all started when I suggested that moving the COG further away from the rear axle would reduce the torque reaction and so reduce front wheel lift.

As moving the COG forwards would also reduce the weight on the rear wheel, it might to be more useful to move the COG upwards. And that this might explain why Clelands ridden out of the saddle climb so well.

67. Originally Posted by james-o
...If you make a great hill-climber for 40% + grades you probably have a bike that won't ride well on other trails...

The bush-whacking do-all in a trials-style ethic of a Cleland is great, but I think it'd need to do so without comproming flow at higher speeds too much to gain popularity. Niche appeal is no bad thing, but marginalisation is a shame...
I share that reservation, but I won't know for sure until I've tried a bike like that. Hill climbing is quite a specialised niche but there are quite a few people who really enjoy it. For example, I place a bike's climbing ability above its descending ability. Tests I have done to improve my lap times showed me that setting the bike up for climbing could save me 3 minutes in a lap, but only cost 1 minute on the descent (obviously that's very subjective and on one course, but it did work for me). The Cleland style bike is just as much a horse for its course as a downhill bike is for its. Maybe we need an uphill riding greasy hill championship

Originally Posted by james-o
...Velobike, I think those old roadsters aren't that far off the geometry of a Jones in some areas, and closer again to many really early MTBs?...
All the old roadsters I have ridden are great on the road, but truly horrible on technical terrain, but naturally I haven't ridden everything. The problem with the old roadsters mainly lies in the front fork rather than the frame geometry, so it's fixable.

Here's some vintage pics of bikes with big wheels:

A design for the future, P437 Cycling 1931

[IMG]

Racing and record breakers:

Coolgardie 1896

Francis Birtles, Warren & Robert Lennie, at Eucla WA, 1907. Lennies attempting Perth-Sydney record

(The last 2 pictures are from the 1980 book,"The Bicycle and the Bush" by Jim Fitzpatrick, which is basically about the use of the bicycle in Australia from the 1880s to early 1900s)

68. Originally Posted by GrahamWallace
Segway weight 83lbs

Segway weight capacity: 260 pound (118 kg) rider and cargo.

That would put the COG of rider and Segway around the top of the riders legs.

HowStuffWorks "Segway: Car Replacement?"
Ok and I think the top of the riders legs are pretty much over the tire contact patch.

69. Originally Posted by Bigwheel
Wow, I am surprised that Cloxxki and David Copperfield have not chimed in on this one.

.
Yes it would be fun to hear from other physics nerds

70. Originally Posted by GrahamWallace
Would then not the simple solution be to fit a short front end and suspension?
Well, there have been short dragsters. I tried to find more info on the length of dragsters but admit I'm not really clear on the reason for the length, but I think it has to do with lateral stability. Since they don't have to corner, there is no limiting length in that respect. Shorter dragsters could have the same downward force on the front (more of the engine weight with less moment arm).

The key distance is to have the COG at the same height as the rear axle as this would cancel out the tendency of the inertia focused at the COG to lift the front. But even then a mechanism is still required to cancel the torque reaction.
You may be right about the force needing to be at the cog to avoid any moment, but if it was, then surely there would be no need to cancel a torque reaction since a force at the COG could cause no rotation at the COG?

71. Originally Posted by Velobike
I very much doubt he needs to chase recognition. He already has that where it matters, except maybe in the small region called USA.

Then why are you guys always coming here to toot you horn? If the us is such a small region then what does it matter what we think?

As for the invention of big wheels, perhaps you could enlighten us as to who you believe had the first application of them to purpose built all terrain bicycles. (And the big wheeled fat tyred bikes of 100 years ago don't count because the geometry of those bikes was road oriented).
To me there was not a true fat tire available for 700c rims until the advent of the "tire" in 1999. Until the you were doing your wandering about on 28" wheels which is not a bad thing, just not what developed in to the real thing.

As I keep trying to point out mtb's are more than about plonking around in the glen. They are an efficient tool that should be able to handle a variety of terrain. Your bikes don't strike me as such but I am glad you all are having fun with them.

72. Originally Posted by GrahamWallace
That's correct.

Newton's Third Law: "To every action there is always an equal and opposite reaction: or the forces of two bodies on each other are always equal and are directed in opposite directions".

This also applies to rotational leverage forces known as torque, where a clockwise torque will always produce an "equal and opposite" anti-clockwise reaction.

If the rear wheel has torque, there must be an equal and opposite force transmitted to the rest of the bike. However, the exact magnitude of this force will depend on the distance away from the axle that you measure it.
I don't get why you are splitting the force out of the torque or what the justification for doing so. Equal and opposite means the torque reaction is the same as the applied torque... whether it is made of a small force and a large arm or vice versa is inconsequential (unless you are building a dragster and the combination of short and heavier is a detriment to acceleration, among other things).
Again, in your own example of COG at infinity the bike tips immediately upon input and you acknowledge this. How is that not a direct contradiction to your argument?
It started for me when you said that raising COG equalizes weight distribution between the wheels, something I still dispute and which is very easily proven false.
I don't think we're going to get anywhere, but it's been a pleasure.

73. Originally Posted by Bigwheel
To me there was not a true fat tire available for 700c rims until the advent of the "tire" in 1999. Until the you were doing your wandering about on 28" wheels which is not a bad thing, just not what developed in to the real thing.

As I keep trying to point out mtb's are more than about plonking around in the glen. They are an efficient tool that should be able to handle a variety of terrain. Your bikes don't strike me as such but I am glad you all are having fun with them.
I think I get what you're saying, if your mtb riding is centred around fast descents then you're right. I don't know about Graham but my riding is on all sorts of terrain, not just "plonking around the glen". Frequently I'll ride 20 or so miles just to get to the trail, come out somewhere on the other side of a mountain and then ride home. Often there will be stretches with no actual trail, just sheep pads or deer tracks, or a scramble over some scree carrying the bike.

For your interest, the 28" wheels with a 2" tyre are the same size as a 29er with a 2.35", but most mountain riding in the UK prior to the mtb era was done on 27" rims, often fitted with 1*1/2" tyres. If you're interested in knowing more about traditional UK offroad riding, look at the the Rough Stuff Fellowship which has been going since 1955.

74. Graham,
I'm starting to see some of what I had missed earlier. I think the term "torque reaction" was throwing me off, but if a force acts away from the center of gravity, it creates a moment.
In my drawing, the front wheel is just starting to lift, so it has a normal force of zero. The rear tire contact patch is the fulcrum around which the COG is rotating.
B is the vertical distance from the COg to the fulcrum and A is the horizontal distance to the fulcrum.
F is the forward driving force from the tire/ground force, which we can say is acting at the COG.
Mg is gravitation force acting on the center of mass, which I have broken into Mgx and Mgy (rather than throwing around sines and cosines )

The sum of moments around the tire contact patch are zero at the time right before front tire lift, so
M clockwise = M counterclockwise

The clockwise moment is:

Mgy (A)

The counterclockwise moment is the sum of the moment caused by the drive force and that of gravity pulling the weight at the COG backward:

Mgx (B) + F(B)

The point of all this is to see what happens if you move the COG upward. distance A stays the same and distance B becomes greater at COG 2. So the counterclockwise moment becomes greater, and the wheel is more likely to lift.

The only way to counter that lift from the higher COG is to also move it forward, so that the clockwise moment again balances the counterclockwise moment.

75. Originally Posted by Bigwheel
To me there was not a true fat tire available for 700c rims until the advent of the "tire" in 1999. Until the you were doing your wandering about on 28" wheels which is not a bad thing, just not what developed in to the real thing.
1999 WTB Nanoraptor dimensions: Otherwise known as "the tire"

700x52c (marked size)
A 700c rim diameter is 622mm add 52mm x 2 for the additional diameter of the tire to get an overall diameter of 726mm or 28.5826771653852 inches.

So it appears that the worlds first 29er tire only measures 29inches when rounded up to the next integer.

In Comparison:
Nokia Hakkapeliitta snow tire dimensions
(The tire used by Geoff Apps on his 1981 700c rimmed Cleland)

700x47c (marked size)
A 700c rim diameter is 622mm add 47mm x 2 for the additional diameter of the tire to get an overall diameter of 716mm or 28.188976377983202 inches.

That makes the Nokia 10mm or 1.4% smaller in diameter

A crucial difference?

From the Cleland viewpoint it means 5mm more mud clearance.

76. Great stuff. Thanks for that bit of history.

77. Originally Posted by smilinsteve
Graham,
I'm starting to see some of what I had missed earlier. I think the term "torque reaction" was throwing me off, but if a force acts away from the center of gravity, it creates a moment.
Hi Steve, I like the drawing, it will definitely help to clarify the nature of the problem. I would have gone onto your technique of taking the sum of the moments acting on the CoG but thought is important to define the nature of torque reactions first. An alternate way of analysing the problem is by working out the sum of the vertical vector components and the horizontal vector components.

The force that acts on the bike below the center of gravity does indeed create a moment but only when drive forces try to accelerate the CoG. These are what I call inertia forces/reaction forces but I believe that they are more commonly called weight transfer or load transfer forces.
Weight transfer - Wikipedia, the free encyclopedia

Weight transfer forces are similar but distinct from torque reaction forces in that they are only present in cases of acceleration or deceleration. Therefore, they do not apply to a Segway when it is climbing at a constant velocity. Torque reaction however, occurs whenever torque is present. For instance if you clamped the rear wheel of an electric bicycle to the floor and then started the motor in a low gear the front wheel would lift into the air even if the bicycle does not move forwards.

Originally Posted by smilinsteve
In my drawing, the front wheel is just starting to lift, so it has a normal force of zero. The rear tire contact patch is the fulcrum around which the COG is rotating.
B is the vertical distance from the COg to the fulcrum and A is the horizontal distance to the fulcrum.
F is the forward driving force from the tire/ground force, which we can say is acting at the COG.
Mg is gravitation force acting on the center of mass, which I have broken into Mgx and Mgy (rather than throwing around sines and cosines )

The sum of moments around the tire contact patch are zero at the time right before front tire lift, so
M clockwise = M counterclockwise

The clockwise moment is:

Mgy (A)

The counterclockwise moment is the sum of the moment caused by the drive force and that of gravity pulling the weight at the COG backward:

Mgx (B) + F(B)

The point of all this is to see what happens if you move the COG upward. distance A stays the same and distance B becomes greater at COG 2. So the counterclockwise moment becomes greater, and the wheel is more likely to lift.

The only way to counter that lift from the higher COG is to also move it forward, so that the clockwise moment again balances the counterclockwise moment.

One issue we need to agree on before we proceed, is the location of the fulcrum that the bike rotates around when the front wheel lifts off the ground. You have Identified it as the rear tire contact patch but may I suggest that it is in fact the rear axle. The reason being is that front lifts off the ground the rear tire does not rotate backwards but caries on moving forward. Wheel axles have a unique property in that unless they fall sideways they cannot be forced nearer to the ground. Also a tire surface rotating makes for a complicated form of fulcrum.

78. Originally Posted by GrahamWallace

One issue we need to agree on before we proceed, is the location of the fulcrum that the bike rotates around when the front wheel lifts off the ground. You have Identified it as the rear tire contact patch but may I suggest that it is in fact the rear axle. The reason being is that front lifts off the ground the rear tire does not rotate backwards but caries on moving forward. Wheel axles have a unique property in that unless they fall sideways they cannot be forced nearer to the ground. Also a tire surface rotating makes for a complicated form of fulcrum.
Yes I was pondering that very point. I have seen the analysis done both ways, and I'm not sure which is more correct. I can see the argument for using the rear axle rather than the tire contact patch as the fulcrum. Doing it this way, I don't think anything changes in the analysis I posted. Distance B becomes shorter (by the radius of the wheel), and distance A stays the same.

79. Originally Posted by meltingfeather
I don't get why you are splitting the force out of the torque or what the justification for doing so. Equal and opposite means the torque reaction is the same as the applied torque...
If the torque is say coming from a motor that is driving a wheel. And the casing of the motor is anchored to a lever, then the measurement of the torque reaction will depend on where along the lever you measure it. Only if you measure it at a point along the lever which is equal to the radius of the wheel will it be of the same magnitude.

[QUOTE=meltingfeather;9737919
Again, in your own example of COG at infinity the bike tips immediately upon input and you acknowledge this. How is that not a direct contradiction to your argument?
[/QUOTE]
I did indeed acknowledge that the inertia of the CoG at infinity would cause the bike to tip.
(See Weight transfer - Wikipedia, the free encyclopedia for the reason why?)
But in theory if the CoG was neither accelerating nor decelerating but traveling at a constant velocity there would be no force present to tip it. Of course traveling at a constant velocity on a bicycle up a steep hill is impossible, because of the pulsed nature of the drivetrain.

[QUOTE=meltingfeather;9737919
It started for me when you said that raising COG equalizes weight distribution between the wheels, something I still dispute and which is very easily proven false.
I don't think we're going to get anywhere, but it's been a pleasure. [/QUOTE]

For my part I am gaining insights and understanding that I would have not without having this discussion.

Namely that what I refer to as CoG inertia more commonly known to motorsport enthusiasts as "axle weight-shift" is not the same phenomenon to torque reaction.

On a bicycle weight shift is only caused by acceleration and deceleration.
"So you can avoid the issues it creates by keeping your speed steady".

Torque reaction is created whenever there is torque present in a system.
It is unavoidable even in constant velocity situations.

Therefore an electric powered mountain bike should be able to climb steeper slopes because of its smooth power delivery and ability to travel with constant velocity.
Moving the center of mass upwards on such a machine would reduce the upwards leverage of the torque reaction trying to lift the front wheel, without raking any weight off the rear wheel.

The next question is could a pedaled bicycle be made that had a torque delivery as smooth as an electric motor? Because if it could be made, such a bike should be able to climb slopes as steep as the Segway.

80. Originally Posted by GrahamWallace
...
One issue we need to agree on before we proceed, is the location of the fulcrum that the bike rotates around when the front wheel lifts off the ground. You have Identified it as the rear tire contact patch but may I suggest that it is in fact the rear axle. The reason being is that front lifts off the ground the rear tire does not rotate backwards but caries on moving forward. Wheel axles have a unique property in that unless they fall sideways they cannot be forced nearer to the ground. Also a tire surface rotating makes for a complicated form of fulcrum.
If I could butt in here with two questions...

What happens when your COG falls in front of the rear axle, but behind the tyre contact patch?

Assuming a constant pedaling speed, what happens to the forward velocity of the rear wheel once the bike starts rotating backwards?

I believe that, in a simple model, the fulcrum is the tyre contact patch.

81. Originally Posted by GrahamWallace
If the torque is say coming from a motor that is driving a wheel. And the casing of the motor is anchored to a lever, then the measurement of the torque reaction will depend on where along the lever you measure it. Only if you measure it at a point along the lever which is equal to the radius of the wheel will it be of the same magnitude.
I get more of what you're saying after thinking about it a bit more.
However, you are saying "equal and opposite" and then turning around and saying the torque reaction depends on distance from the pivot. It does not. Equal and opposite is valid, the mechanism by which the force component of the torque reaction is reduced is the lengthening of the lever arm, which multiplies the force from a torque perspective. Since the mass stays the same in this case, I can see what you are talking about, but think the effect is negligible. You have been talking about torque reactions at constant velocity, which requires only enough torque to counter drag. Weight shift, especially on grade and with a pulsed drive, I think is much more of a driver, and raising the COG is detrimental in that case.
Originally Posted by GrahamWallace
For my part I am gaining insights and understanding that I would have not without having this discussion.
Me too. I thought we had reached an impasse, which I didn't understand, since you seem to be a logical and educated fellow, so I gave it a bit more thought and I think I was looking past torque reactions to what I think is more of an effect.

Originally Posted by GrahamWallace
Namely that what I refer to as CoG inertia more commonly known to motorsport enthusiasts as "axle weight-shift" is not the same phenomenon to torque reaction.
I get it now.

Originally Posted by GrahamWallace
On a bicycle weight shift is only caused by acceleration and deceleration.
"So you can avoid the issues it creates by keeping your speed steady".
It is also caused by grade. Changing the angle of incline of the bike has the same effect as summing the vertical weight vector with the horizontal one that results from acceleration on level ground.

Originally Posted by GrahamWallace
Torque reaction is created whenever there is torque present in a system.
It is unavoidable even in constant velocity situations.
I agree.

Originally Posted by GrahamWallace
Therefore an electric powered mountain bike should be able to climb steeper slopes because of its smooth power delivery and ability to travel with constant velocity.
Moving the center of mass upwards on such a machine would reduce the upwards leverage of the torque reaction trying to lift the front wheel, without raking any weight off the rear wheel.
This is impossible. The sum of the ground reactions at the two contact patches must equal the mass of the bike+rider.
Like a weight distributing trailer hitch, the introduction of the moment causes a skew in the distribution, but not a difference in the sum of the ground reactions.
Originally Posted by GrahamWallace
The next question is could a pedaled bicycle be made that had a torque delivery as smooth as an electric motor? Because if it could be made, such a bike should be able to climb slopes as steep as the Segway.
Sure... pedal it with a robot.

82. Wow the numbers and calculations in this thread are too much. You can spin it any way you want but the modern 700c Mt. Bike was based on the advent of the "tire". It created a market for high volume, low pressure tires that has gained an enormous following, won World Cups and Olympic Gold Medals as well as use on thousands of bikes that just plain old people ride in all sorts of ways and places all over the world. Not just some boggy glen in the British Isles.

Here is a photo of a bike with Hakkapaleta's on it:

While it may be close to the same height it lacks the volume and that folks is what made me subscribe back in 99'.

For some reason my iPad won't allow me to post multiple pics as I was going to insert one of a "tire" for reference but just go look at what you have on your bike. Not a huge difference but big enough to make a difference.

I would venture to say that the new Knard tire is an equal step above and I can't wait to give them a go. Life is incremental and that is what keeps it interesting. Innovation is key, not mucking about trying to figure out which came first, the chicken or the egg.

83. Originally Posted by sbitw
If I could butt in here with two questions...

What happens when your COG falls in front of the rear axle, but behind the tyre contact patch?
Hi sbitw,

That's an interesting question. In my logic it depends where you believe the fulcrum is. If it is the axle then the CoG is still creating a small downwards moment of rotation even though it is directly above the contact point. If the contact point is the fulcrum then there will be no moment of rotation so the front wheel must lift. This will happen even when the bike is moving forward at a constant speed and there is no forces pushing the CoG backwards as the torque reaction alone will lift the front wheel. The front wheel accelerates as it lifts, the energy required for this is stolen from the torque of the rear wheel. But not enough energy will be required to stop the rear wheel from rotating forwards. If pedaling continues then the bike will continue to accelerate in its rotation around the axle but with gravity now assisting the back flip where earlier on it was resisting it.

For the rear contact patch to be acting as the fulcrum the wheel would have to rotate backwards whilst its brake was applied.

Originally Posted by sbitw
Assuming a constant pedaling speed, what happens to the forward velocity of the rear wheel once the bike starts rotating backwards?
The torque of the rear will be split between moving the bike forward and lifting the front end so the rear wheel must lose some forwards velocity.

Originally Posted by sbitw
I believe that, in a simple model, the fulcrum is the tyre contact patch.
There may be an error in my reasoning,
If there is, I am hoping that someone will spot it and point it out.

84. Originally Posted by Bigwheel
Tucson Mountain Park?

85. Originally Posted by GrahamWallace
1999 WTB Nanoraptor dimensions: Otherwise known as "the tire"

700x52c (marked size)
A 700c rim diameter is 622mm add 52mm x 2 for the additional diameter of the tire to get an overall diameter of 726mm or 28.5826771653852 inches.

So it appears that the worlds first 29er tire only measures 29inches when rounded up to the next integer.

In Comparison:
Nokia Hakkapeliitta snow tire dimensions
(The tire used by Geoff Apps on his 1981 700c rimmed Cleland)

700x47c (marked size)
A 700c rim diameter is 622mm add 47mm x 2 for the additional diameter of the tire to get an overall diameter of 716mm or 28.188976377983202 inches.

That makes the Nokia 10mm or 1.4% smaller in diameter

A crucial difference?

From the Cleland viewpoint it means 5mm more mud clearance.

Originally Posted by Bigwheel
WYou can spin it any way you want but the modern 700c Mt. Bike was based on the advent of the "tire". It created a market for high volume, low pressure tires that has gained an enormous following, won World Cups and Olympic Gold Medals as well as use on thousands of bikes that just plain old people ride in all sorts of ways and places all over the world. Not just some boggy glen in the British Isles.
But there were a string of events that led to the "tire",

Originally Posted by Bigwheel
While it may be close to the same height it lacks the volume and that folks is what made me subscribe back in 99'.
I agree.

Originally Posted by Bigwheel
For some reason my iPad won't allow me to post multiple pics as I was going to insert one of a "tire" for reference but just go look at what you have on your bike. Not a huge difference but big enough to make a difference.
The picture you show is not a 700x47c Nokia Hakkaoeliitta. Here is picture of one like those Geoff Apps exported to the US in the 1980's:

And Here is a picture of a 1988 Bruce Gordon 700c "Rock N Road" tire. The tire that inspired Wes Williams, who with Garry Fisher commissioned the Nano-raptor. Notice that the tread pattern was copied from the Hakkapeliitta.

86. Originally Posted by GrahamWallace
The force that acts on the bike below the center of gravity does indeed create a moment but only when drive forces try to accelerate the CoG. These are what I call inertia forces/reaction forces but I believe that they are more commonly called weight transfer or load transfer forces.
Weight transfer - Wikipedia, the free encyclopedia
I think the torque reaction you keep talking about is the moment caused by the force pushing the bike forward. At constant velocity, this force is balanced with gravity and friction holding it back. Balanced forces, no net moment. When accelerating this force is larger than the counteracting forces. This creates a total moment about the COG. I don't think there is any other torque.

Weight transfer forces are similar but distinct from torque reaction forces in that they are only present in cases of acceleration or deceleration. Therefore, they do not apply to a Segway when it is climbing at a constant velocity. Torque reaction however, occurs whenever torque is present. For instance if you clamped the rear wheel of an electric bicycle to the floor and then started the motor in a low gear the front wheel would lift into the air even if the bicycle does not move forwards.
I can't picture the configuration of your electric bike, but if I clamp the rear wheel of a regular bike and step on the crank, the front end will not rise.

Why does the bike have to be electric, and how does that change things? If an electric motor shaft is attached directly to the bottom bracket spindle, motor torque is converted to chain tension. Chain tension is converted to force on the rear cog. Since the wheel is clamped, there is no motion. No torque can be transmitted from the wheel to frame because the wheel is attached to the frame only by the axle with ball bearings. Only linear force can transfer from wheel to frame. You can't transfer a torque through a pivot point.

When I went to school, the free body diagram would identify all forces and moments in the system. This torque reaction you are talking about seems to be something not identified in the drawing I made? Then where does it come from, where is it applied, and where is the force that generates it?

87. Originally Posted by Originally Posted by GrahamWallace
Torque reaction is created whenever there is torque present in a system.
It is unavoidable even in constant velocity situations.
Originally Posted by Meltingfeather
I agree.

I disagree. Is this the heart of the matter here?

Newtons third law does not state that every torque creates an equal and opposite torque.

Torque is just a force that creates rotational motion. That force could have a reaction that creates no rotational motion.

Example:

Force from your leg > torque at bottom bracket > Tension on chain > torque at wheel hub > force from tire on ground > equal and opposite reaction is linear force on wheel axle pushing bike forward.

See how forces convert to torques which convert back to linear force?
Forces have reactions. Forces can cause torque, or not. Torque can be converted to linear force with no equal and opposite torque.

Example 2:

Take your bike as above. Pin one end of the chain to your chainring. The other end of the chain is not attached to your wheel, but to the center of a wooden box sitting on the road.
Push down on the crank;

Force from leg > torque on bottom bracket > tension in chain > Linear motion of box.

Example 3

Same as example 2 but chain is attached to the top of a tall skinny box with high coefficient of friction between the box and the ground;

Force from leg > torque on BB > tension on chain > torque on box, causing it to fall over.

The difference between example 2 and 3 is whether or not the force from the chain acts through the COG. In example 3 it doesn't, and creates a torque.

I guess what I'm getting at here Graham, is that if there is a "torque reaction" at the center of gravity of the rider, then it is caused by some force acting a distance away from the COG.

I see the torque from pedaling creating a ground force below the COG that creates a moment about it.
You seem to think there is another source for torque around the COG, but I don't see it.

88. Originally Posted by smilinsteve
I think the torque reaction you keep talking about is the moment caused by the force pushing the bike forward. At constant velocity, this force is balanced with gravity and friction holding it back. Balanced forces, no net moment. When accelerating this force is larger than the counteracting forces. This creates a total moment about the COG. I don't think there is any other torque.

Anything that rotates clockwise will create an anti-clockwise reaction. If you sit in a revolving chair, hold your arms outwards and then move them quickly clockwise, your body and the chair will rotate anti-clockwise as a reaction to the torque generated by the moving arms. Likewise if a bicycle wheel rotates clockwise there will be an anti-clockwise reaction in the frame.

Originally Posted by smilinsteve
I can't picture the configuration of your electric bike, but if I clamp the rear wheel of a regular bike and step on the crank, the front end will not rise.
It may if you put it in a low gear, and definitely would if you placed the bike on a steep incline.

Originally Posted by smilinsteve
Why does the bike have to be electric, and how does that change things? If an electric motor shaft is attached directly to the bottom bracket spindle, motor torque is converted to chain tension. Chain tension is converted to force on the rear cog. Since the wheel is clamped, there is no motion. No torque can be transmitted from the wheel to frame because the wheel is attached to the frame only by the axle with ball bearings. Only linear force can transfer from wheel to frame. You can't transfer a torque through a pivot point.
There is also an axle and ball bearing connection between a revolving chair and its base but it can still be made to rotate without you touching anything around you.

Originally Posted by smilinsteve
When I went to school, the free body diagram would identify all forces and moments in the system. This torque reaction you are talking about seems to be something not identified in the drawing I made? Then where does it come from, where is it applied, and where is the force that generates it?
1/ It comes from the rear wheel torque.
2/ It is applied to anything that stops it from moving freely.
3/ My legs and feet and then the mechanical advantage is increased via low gearing.

89. Originally Posted by smilinsteve
I disagree. Is this the heart of the matter here?
I'm not sure.
Originally Posted by smilinsteve
Newtons third law does not state that every torque creates an equal and opposite torque.
But they do. This is the basis for the sum of moments analytical tool in statics.

Originally Posted by smilinsteve
Torque is just a force that creates rotational motion. That force could have a reaction that creates no rotational motion.
I do not believe it can.

Originally Posted by smilinsteve
Example:

Force from your leg > torque at bottom bracket > Tension on chain > torque at wheel hub > force from tire on ground > equal and opposite reaction is linear force on wheel axle pushing bike forward.

See how forces convert to torques which convert back to linear force?
Forces have reactions. Forces can cause torque, or not. Torque can be converted to linear force with no equal and opposite torque.
You took one force and translated it through at least four different bodies to arrive at your conclusion, which I'd say is a little larger in scope than what Newton intended.
The reactions are between the individual bodies, not translated through a machine. The equal and opposite force from tire on ground is ground on tire. Likewise for every piece in your example.

90. I will try and explain torque reaction a different way.

If you constrain a large force it will try and find anyway it can to break free.

If you constrain the torque force in a rear wheel by giving it a large amount of resistance from the ground, then it will look for other places to escape. It could break the chain for instance, or it could cause the front wheel to lift off the ground. And the bicycle doesn't even need to move for this to happen.

Newtons third Law refers to forces. All forces.

They can be restrained and not visible like the tension in a chain.
The same is true for reaction forces. But sometimes they find ways to unexpectedly break free.

If torque reaction does not exist why invent "Torque Reaction Arms" ?

BMF Torque Wrench Co. - Model SD Wrenches

91. Originally Posted by meltingfeather
You took one force and translated it through at least four different bodies to arrive at your conclusion, which I'd say is a little larger in scope than what Newton intended.
The reactions are between the individual bodies, not translated through a machine. The equal and opposite force from tire on ground is ground on tire. Likewise for every piece in your example.
Well, I'm thinking the action reaction pairs newton talked about are forces. Forces can create torque or not. So in my example, how would you break down the action reaction pairs?

Force on pedal creates an opposite force from pedal to foot!
In this case, it doesn't matter if that force from foot to pedal creates a torque or doesn't, but it does.

In my example of the chain attached to the top of the box, the reaction to the force on the box by the chain is the force on the chain by the box. the chain creates a moment that topples the box, but what is the "torque reaction?" there doesn't need to be an equal and opposite torque, only an equal and opposite force.

92. Originally Posted by smilinsteve
Newtons third law does not state that every torque creates an equal and opposite torque.
Originally Posted by meltingfeather
But they do. This is the basis for the sum of moments analytical tool in statics.
In statics, if a body isn't rotating, you know there is no net moment on it.

But what does that have to do with bodys that are rotating? If I am in outer space and push on one end of an asteroid floting freely, the equal and opposite reaction to my push on the asteroid is a force against my hand. So I move backwards, and the asteroid starts rotating. So my push created a torque, but the equal and opposite reaction was a linear force against me causing me to move, not rotate.

93. This is going to be my last post to you Graham but let me say this. Stick to the things you know about what led Geoff to using 700c "off trail" and don't pretend to know what led Wes to his part in the "tire" or Gary Fisher either because you obviously don't.

And btw, that pic of the tacoed wheel in your reply to my last post was probably not the best way to illustrate your point.

94. Originally Posted by smilinsteve
In statics, if a body isn't rotating, you know there is no net moment on it.

But what does that have to do with bodys that are rotating? If I am in outer space and push on one end of an asteroid floting freely, the equal and opposite reaction to my push on the asteroid is a force against my hand. So I move backwards, and the asteroid starts rotating. So my push created a torque, but the equal and opposite reaction was a linear force against me causing me to move, not rotate.
You could also push the asteroid in away that makes you rotate.It depends on where in your body the push originates relative to your center of mass.

95. There is also an axle and ball bearing connection between a revolving chair and its base but it can still be made to rotate without you touching anything around you
In your swivel chair example the chair is analagous to the wheel and the base of the chair is analagous to the frame. You can spin that chair all you want, but you aren't going to transmit any torque to the base.

It may if you put it in a low gear, and definitely would if you placed the bike on a steep incline.
Your bike with the clamped wheel is an easy experiment to do at home, so lets do that and see what happens. I am sure you can not lift the front end of the bike. Draw a free body diagram and show the force that lifts the bike. ?
If the rear wheel is clamped, say by a rear disc brake, the force on the crank creates internal forces in the frame transmitted from crank to chain to axle, and from brake rotor to caliper to frame, etc. There is no force between the bike and ground.
Internal forces can not cause motion. Its like trying to grab the arms of your chair and lifting yourself off the ground while you are sitting it

96. If torque reaction does not exist why invent "Torque Reaction Arms" ?
I already mentioned nut drivers as one example of a torque reaction. I don't see the analogy to a rider pedaling uphill.

Again, a free body diagram, like I posted, is the way you analyse the forces and moments in a system.
The way to do this problem is to evaluate the moments at the point of tire lift. I already showed the effect of moving the COG upward.

97. Originally Posted by Bigwheel
This is going to be my last post to you Graham but let me say this. Stick to the things you know about what led Geoff to using 700c "off trail" and don't pretend to know what led Wes to his part in the "tire" or Gary Fisher either because you obviously don't.
But I would like to know where I have gone wrong?

Originally Posted by Bigwheel
And btw, that pic of the tacoed wheel in your reply to my last post was probably not the best way to illustrate your point.
Is a prototype that is 30years before its time not allowed a few teething troubles?

Here is an independent view of 29er history:
Guitar Ted Productions: The Beginnings Of The Modern 29"er: A History

98. Originally Posted by GrahamWallace
You could also push the asteroid in away that makes you rotate.It depends on where in your body the push originates relative to your center of mass.
Agreed, but I am trying to disprove what you said, that any system with a torque must have a torque reaction.

99. Originally Posted by smilinsteve
Agreed, but I am trying to disprove what you said, that any system with a torque must have a torque reaction.
And you have, and are right in pointing out that a force can be turned into a torque and a torque into a force.

Now you need to find a stationary swivel chair, lift your feet off the floor, introduce some torque by waving your arms and legs and see if the chair rotates.

Or failing that explain why a hovering helicopter needs a tail rotor?

Or failing that research the difference between a torque reaction arm and a nut driver.

Or drill a hole in a wall and see if the electric drill tries to rotate the opposite way to the bit.

If Newton had meant that only a linear force has an equal and opposite reaction. Would not he have said so?

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