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  1. #201
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    Quote Originally Posted by GrahamWallace View Post
    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?
    Other than the chair, the examples are all related to shaft torque and when I google torque reaction, most hits also refer to shafts. These are obvious examples of torque reaction.
    For the bike riding up the hill, you need to draw a FBD with all forces and calculate all moments. I still think that my FBD shows all forces and moments.

    If you turn your bike upside down and turn the cranks, the cranks turn the wheel. The force of your arm applies a torque to the hub of the wheel causing it to accelerate.
    What other forces are acting on the bike?

    As for the chair, I am flapping away furiously, and not getting much movement, but want to think a bit more about this one, and hopefully no one will walk into my office.

  2. #202
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    Quote Originally Posted by smilinsteve View Post
    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 would also rotate, unless you just happened to push with a force that intersected your COG.
    Newton's 3rd law stops at the equal and opposite reactions... what happens to each body after that is 2nd law stuff (F=ma, or in the case of rotation, τ=Iα)
    Quote Originally Posted by pvd
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  3. #203
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    Quote Originally Posted by smilinsteve View Post
    For the bike riding up the hill, you need to draw a FBD with all forces and calculate all moments. I still think that my FBD shows all forces and moments.
    I think the drive force needs to be shown at the contact patch. It does not act at the COG, and that may be the issue.
    Quote Originally Posted by smilinsteve View Post
    If you turn your bike upside down and turn the cranks, the cranks turn the wheel. The force of your arm applies a torque to the hub of the wheel causing it to accelerate.
    What other forces are acting on the bike?
    The inertia of the wheel pushes back against the torque of the driving force.
    Quote Originally Posted by smilinsteve View Post
    As for the chair, I am flapping away furiously, and not getting much movement, but want to think a bit more about this one, and hopefully no one will walk into my office.
    You will only get small, local rotation, because in that case, the only "outside" force acting is the friction in the bearings/rotating mechanism of the chair, which is small. Without outside force there can be no net movement. Do "The Twist" and your chair should move back and forth opposite your torso.
    Quote Originally Posted by pvd
    Time to stop believing the hype and start doing some science.
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  4. #204
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    Quote Originally Posted by smilinsteve View Post
    ...As for the chair, I am flapping away furiously, and not getting much movement, but want to think a bit more about this one, and hopefully no one will walk into my office.
    Just think, if only Newton had an electric drill, all this would have been worked out ages ago...

    Bigwheel, I don't know why you seem so upset about the claim to be first with the large wheel offroad bike. It doesn't take anything away from the guys who popularised/commercialised it or indeed from the USA mtb industry without whom we wouldn't have the modern mtb.

    The likes of the British bike industry studiously ignored the fact that there were people who spent their time riding offroad, and instead of leading the way continued to produce nothing but roadbikes. The demand had existed since the dawn of the bicycle age, and they constantly remained blind to it.

    I have a collection of bike books and magazines starting from the 1870s, and one constant in them is articles on pass-storming as it was called then. So the demand was there for suitable tyres and frames, which is possibly a chicken and egg thing.

    And without the Hakka, there would have been no big wheel mtb, so we should also be thanking the Finns.
    As little bike as possible, as silent as possible.
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  5. #205
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    Quote Originally Posted by smilinsteve View Post
    Other than the chair, the examples are all related to shaft torque and when I google torque reaction, most hits also refer to shafts. These are obvious examples of torque reaction.

    Could shaft be the same as = Axle?


    Quote Originally Posted by smilinsteve View Post
    I still think that my FBD shows all forces and moments.
    The big omission is the rear wheel's coefficient of friction with the ground. If the rear wheel can't slip the CoG can move right to the front of the wheel-base without any problem.

    For a FBD It would probably be a good idea to place the CoG directly above the center of the wheelbase so that the weight is split between the wheels.

    Quote Originally Posted by smilinsteve View Post
    For the bike riding up the hill, you need to draw a FBD with all forces and calculate all moments. I still think that my FBD shows all forces and moments.
    The more I think of this the more I think it may be possible. It would take some work and there would be two versions. One for an accelerating CoG. And one for a constant velocity CoG. (Someone mentioned earlier that a CoG rising up an incline whilst being pulled backwards by gravity is accelerating? Surely not as this just requires the motive vector to be stronger than the rearward pull of Gravity.

    Quote Originally Posted by smilinsteve View Post
    If you turn your bike upside down and turn the cranks, the cranks turn the wheel. The force of your arm applies a torque to the hub of the wheel causing it to accelerate.
    What other forces are acting on the bike?
    There is a small torque reaction force created by the rotational inertia of the mass off the rotating wheel though this is much larger during acceleration and deceleration. for a big torque reaction the wheel needs to be experiencing drag. In the case of a helicopter the drag comes the air resistance of the rotor blades.

    Quote Originally Posted by smilinsteve View Post
    As for the chair, I am flapping away furiously, and not getting much movement, but want to think a bit more about this one, and hopefully no one will walk into my office.
    Well I'am sitting here in my swivel chair spinning happily. Good job I am not at work!
    Physics Swivel Chair Olympics vol 7 - YouTube

    If you take a top view helicopter with its rotors rotating clockwise and its tail rotor to the right. Is this not similar to a bicycle when viewed from the side with the rear wheel representing the main rotors, the tail representing the top tube and the rotor now pushing downwards? The main difference being that in the bicycle it is rear wheel drag not air resistance causing a torque reaction. And that the down force of the tail rotor is replaced by the weight of the CoG in the bike?
    Last edited by GrahamWallace; 10-02-2012 at 03:56 PM. Reason: clarification

  6. #206
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    Pretty long thread for just five guys talking to eachother. Interesting topic though. Love the Fox hunting outfits you British guys wear!

  7. #207
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    Quote Originally Posted by ghettocop View Post
    Pretty long thread for just five guys talking to eachother. Interesting topic though.
    If we can get past the question of does torque reaction exist? And is it relevant issue? It may get even more Interesting

    Quote Originally Posted by ghettocop View Post
    Love the Fox hunting outfits you British guys wear!
    Thanks! The real reason that our bikes have high handlebars is so we can hold our umbrellas when we ride in the rain Though they also make a good rest for the shotgun.
    Last edited by GrahamWallace; 10-03-2012 at 02:21 PM. Reason: Bad punctuation

  8. #208
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    Yes a long line of people who contributed. Some credited for their input, some not., And some claiming they got the whole Idea just by looking at the size of the rocks on the Pearl Pass.

  9. #209
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    Quote Originally Posted by GrahamWallace View Post
    The more I think of this the more I think it may be possible. It would take some work and there would be two versions. One for an accelerating CoG. And one for a constant velocity CoG.
    One version is fine. It is either a statics problem (forces and moments balance), or a dynamics problem (forces and moments are partially balanced by masses accelerating).

    Quote Originally Posted by GrahamWallace View Post
    (Someone mentioned earlier that a CoG rising up an incline whilst being pulled backwards by gravity is accelerating? Surely not as this just requires the motive vector to be stronger than the rearward pull of Gravity.
    I don't think this is referencing a comment I made, because I did not say this, but just to be sure, because I did make a statement about "net effect" of an incline, what I said was that the incline, which results in the weight vector of the COG pointing partially rearward (toward the pivot), has the same net effect on the force balance as combining a vertical weight vector with a horizontal force vector accelerating the mass-- a rearward-pointing vector at the COG, which will tip the bike, given the proper geometry and magnitudes.
    Quote Originally Posted by pvd
    Time to stop believing the hype and start doing some science.
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  10. #210
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    Quote Originally Posted by meltingfeather View Post
    you would also rotate, unless you just happened to push with a force that intersected your COG.
    Newton's 3rd law stops at the equal and opposite reactions... what happens to each body after that is 2nd law stuff (F=ma, or in the case of rotation, τ=Iα)
    Yes! That's pretty much what I was getting at. A force creates a reaction that may or may not create a moment. If it acts through the COG, then there is no moment. Any moments in the bicycle free body diagram are caused by forces acting away from the COG. If there is any torque reaction to be considered, then the force and the point of application needs to be identified.

  11. #211
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    Quote Originally Posted by meltingfeather View Post
    I think the drive force needs to be shown at the contact patch. It does not act at the COG, and that may be the issue.
    That took some thought. I think you are right, but I have seen that it is typical convention to show forces acting at the COG. The driving force at the contact patch pushes the bike forward, which creates a counterclockwise moment at the COG. So, if you show the force at the COG, it is pointing backwards to create the counterclockwise moment.

  12. #212
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    Quote Originally Posted by woodhzak View Post
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    I can only hope I am making a bit more sense than woodhzak

  13. #213
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    Here's a discussion about a physics problem: Calculate the rate of acceleration to cause a motorcycle to wheelie. It's a forum, so you have to wade through it, but the bottom line is the wheelie is caused by the drive force creating a moment on the COG until it overcomes the downward force of weight on the front wheel.

    Motorcycle Wheelie (Angular Momentum)

  14. #214
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    Quote Originally Posted by smilinsteve View Post
    Here's a discussion about a physics problem: Calculate the rate of acceleration to cause a motorcycle to wheelie. It's a forum, so you have to wade through it, but the bottom line is the wheelie is caused by the drive force creating a moment on the COG until it overcomes the downward force of weight on the front wheel.

    Motorcycle Wheelie (Angular Momentum)
    Yes. The "torque reaction" Mr. Wallace speaks of is the appropriate component of the weight vector balancing against the rear wheel moment. Once the moment has overcome the weight distributed to the front wheel (by putting it on the rear wheel), additional moment causes the front wheel to lift.
    The bike being on an incline gets you part of the way there by distributing weight to the rear wheel as a result of the geometry.
    Quote Originally Posted by pvd
    Time to stop believing the hype and start doing some science.
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  15. #215
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    Wow. This thread was so interesting to begin with. Now it is mostly confusing.

    Back on the subject of bikes, I have another idea for Geoff and Graham to consider. Its called fat front or half fat. looking at the video of the riding in the stream, I noticed a lot swaying of handle bars. Its the same in the picture too. Being a fat bike rider myself, I can say that the fat tires hold a straighter line while riding over slippery rocks. With the availability of fat bike parts being abundant, it would be easy to convert a Cleland to fat front.

    Here is a thread with pics:
    Half the Fat?

    Just something to consider.

  16. #216
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    Quote Originally Posted by GrahamWallace View Post
    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.



    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.



    There may be an error in my reasoning,
    If there is, I am hoping that someone will spot it and point it out.
    In the process of responding to you, I've talked myself out of my own position. I now agree that the bike rotates about the rear axle.

    In the case where the COG is in front of the axle and behind the contact patch (assuming constant cadence and no wheel slip), the bike does rotate backwards about the rear axle; the rotation is due to the force applied tangentially to the wheel at the contact patch.

    I'll butt out now.

    sbitw

  17. #217
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    Quote Originally Posted by sagealmighty View Post
    Wow. This thread was so interesting to begin with. Now it is mostly confusing.

    Back on the subject of bikes, I have another idea for Geoff and Graham to consider. Its called fat front or half fat. looking at the video of the riding in the stream, I noticed a lot swaying of handle bars. Its the same in the picture too. Being a fat bike rider myself, I can say that the fat tires hold a straighter line while riding over slippery rocks. With the availability of fat bike parts being abundant, it would be easy to convert a Cleland to fat front.

    Here is a thread with pics:
    Half the Fat?

    Just something to consider.
    Yes, that's right. In the video I was using 2.4" Schwalbe Racing Ralphs as an experiment with lighter tyres. As you can see, the steering is made less controllable running them at the low pressures necessary for grip and shock absorbtion. This seems mainly to be due to the thin sidewalls of the Ralphs.
    I've now fitted 2.5" WTB Dissents again, heavier tyre with thicker sidewalls, much better.
    My next move, when I can afford it, is to fit 2.75" tyres which have very stiff sidewalls and will require little or no air inside.
    I'm reluctant to go down the fat bike route for the time being until I have had a chance to try one out. That course of design endeavour will also require a new frame and lots of other stuff, so no action until I have a lot more money to hand (probably never!).

  18. #218
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    Quote Originally Posted by sagealmighty View Post
    Wow. This thread was so interesting to begin with. Now it is mostly confusing.

    > SNIP <

    Just something to consider.
    Yes, An interesting thread! I agree that it has become confusing to a bit but still interesting.
    If I may add and I will keep it simple....

    A static bike is one that has no motion or forces on it.... Once you start moving all thing change with many dynamic forces that come into play.
    Inclines, declines, obstacles, braking and pedaling all are dynamic forces, with gravity working all the time!
    Forward motion literaly starts the wheels turning. Once the wheels are rotating, gyroscopic action comes into play, using this gyro-force to your advantage can make the ride. We all started out with no intuition of how to ride a bike, usually we started with training wheels till we learn and feel comfortable with the "feel" of the dynamic bike.
    The old adage that you never forget how to ride a bike is true! but we learn more as we ride more.... for some it is a marvel how they can do these things, like Danny MacAskill! What he does is nothing short of amazing!
    I think it has a lot to do with your inner ear and it's balance system, with all these math formulas with this thread, it still comes down to skill, experience and good luck.... a flat going through a course is BAD luck!
    With the gyro-action of the bikes wheels and making use of this stabilizing force, a rider can maintain his course well using the rotating wheels to his advantage. Using the CG of the bike rider is mostly a sense of balance and becomes intuitive with practice and skill.
    Here is a link with better understanding of bike wheels in motion, gyros....

    Bicycle Wheel Gyro: Mechanics Science Project | Exploratorium Science Snacks
    Attached Images Attached Images  
    "I don't ride a bike, I'm the pilot!"

  19. #219
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    Quote Originally Posted by smilinsteve View Post
    Other than the chair, the examples are all related to shaft torque and when I google torque reaction, most hits also refer to shafts. These are obvious examples of torque reaction.
    For the bike riding up the hill, you need to draw a FBD with all forces and calculate all moments. I still think that my FBD shows all forces and moments.

    If you turn your bike upside down and turn the cranks, the cranks turn the wheel. The force of your arm applies a torque to the hub of the wheel causing it to accelerate.
    What other forces are acting on the bike?

    As for the chair, I am flapping away furiously, and not getting much movement, but want to think a bit more about this one, and hopefully no one will walk into my office.
    Lol!

    poikaa
    "I don't ride a bike, I'm the pilot!"

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    Danny MacAskill video

    Just in case you have not seen Danny in action....

    Inspired Bicycles - Danny MacAskill April 2009 - YouTube

    How this applies to mountain biking? I suppose it does!

    poikaa
    "I don't ride a bike, I'm the pilot!"

  21. #221
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    If we can get past the question of does torque reaction exist? And is it relevant issue?
    q1 - I think yes, it's part of the forces you feel lifting the front as you pop a wheelie. Weight shift + torque reaction in balance - at that moment the lead foot pushes down, eight moves back and the acceleration / torque and rearward weight shift pop the front up. On a climb, your weight on the front wheel / COG overcomes the pop and you move forwards. thf Q2 - I think no, it's a smaller component in a larger force system if you climb in 'normal' balance.
    With COG at infinite height above but same position between the wheelbase, the bike wheelies again, I see the points made there. But If the COG is 'tied' to the bike, ie muscles acting in arms and legs, maybe not - muscle input mean the forces here are too complex for diagrams.

    At climbing gradient limits, torque resulting in wheelspin, or positioning on the bike compromising power output and/or balance is the issue. Or the wet/rocky/ etc terrain.
    It's all too variable and complex to model imo, but I appreciate the force diagram points as it's food for thought (and some confusion, followed by 'maybe I'll just go for a ride'!)

    Sorry If I'm simplifying. Just coming back to this after thinking about what limits climbing after a couple of rides on steeper hills. Also knowing that I'm less physicist and more ride-by-feel type )

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    Quote Originally Posted by smilinsteve View Post
    Here's a discussion about a physics problem: Calculate the rate of acceleration to cause a motorcycle to wheelie. It's a forum, so you have to wade through it, but the bottom line is the wheelie is caused by the drive force creating a moment on the COG until it overcomes the downward force of weight on the front wheel.

    Motorcycle Wheelie (Angular Momentum)
    Quote Originally Posted by meltingfeather View Post
    Yes. The "torque reaction" Mr. Wallace speaks of is the appropriate component of the weight vector balancing against the rear wheel moment. Once the moment has overcome the weight distributed to the front wheel (by putting it on the rear wheel), additional moment causes the front wheel to lift.
    The bike being on an incline gets you part of the way there by distributing weight to the rear wheel as a result of the geometry.
    Does this mean that at the point of front wheel lift, the rider moves his CoG further away from the rear axle, the liftoff would be prevented?

    And that this would happen even if the COG was only moved upwards?

    My thinking here is that a vertical movement of the CoG would reduce the effect of the rear wheel moment and so take the system out of equilibrium.

    (These are based on the precept that the rear wheel torque moment is constant so the motorbike is not accelerating).
    Last edited by GrahamWallace; 10-04-2012 at 03:15 PM. Reason: clarification

  23. #223
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    Quote Originally Posted by GrahamWallace View Post
    Does this mean that at the point of front wheel lift, the rider moves his CoG further away from the rear axle, the liftoff would be prevented?

    And that this would happen even if the COG was only moved upwards?

    My thinking here is that a vertical movement of the CoG would reduce the effect of the rear wheel moment and so take the system out of equilibrium.

    (These are based on the precept that the rear wheel torque moment is constant so the motorbike is not accelerating).
    No and no.
    The moment is countering the weight vector, specifically the component of the weight vector that is perpendicular to the moment arm (wheel radius), which does not change when the COG moves upward on level ground. It shrinks as the COG moves "upward" on a bike on an incline.
    If the COG moves forward then the moment arm increases.
    (ignore the sub D in the second force equation)
    Attached Thumbnails Attached Thumbnails Cleland: The original big wheeled off-road bicycle?-fbd-tip.jpg  

    Last edited by meltingfeather; 10-04-2012 at 08:53 PM.
    Quote Originally Posted by pvd
    Time to stop believing the hype and start doing some science.
    29er Tire Weight Database

  24. #224
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    Just thinking about climbing a slope:
    Assuming the bicycle remains perfectly vertical, the fulcrum about which the bicycle is rotating is the rear wheel axle. Even if a wheelie does not occur, the pressure applied to the front wheel by way of rider weight transfer, is counteracting the tendency for the front wheel to lift, so there is always a potential wheelie.
    However, the bicycle does not always remain perfectly vertical, so the rider is simultaniously still conteracting the tendancy for the bicycle to fall sideways. The fulcrum for this conteraction is the contact patch of the rear wheel.
    Therefore, one must consider whether both these counteractions can be separated, or if one contributes to the other, if they interact or if in reality they can never be separated.
    Last edited by GeoffApps; 10-05-2012 at 02:14 AM. Reason: clarification

  25. #225
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    Quote Originally Posted by GrahamWallace View Post
    Does this mean that at the point of front wheel lift, the rider moves his CoG further away from the rear axle, the liftoff would be prevented?

    And that this would happen even if the COG was only moved upwards?

    My thinking here is that a vertical movement of the CoG would reduce the effect of the rear wheel moment and so take the system out of equilibrium.

    (These are based on the precept that the rear wheel torque moment is constant so the motorbike is not accelerating).
    I think a key factor missing from this perspective is the freedom of the pivot to move.
    As we have acknowledged, with the COG at infinity, the COG does not move, yet the bike tips because there is relative rotation of COG about the axle (due to movement of the pivot, not the COG).
    As the COG moves away from the axle and experiences less acceleration due to the torque reaction, the acceleration of the pivot point out from under the COG increases.
    The relative motion is what's missing from the explanations I'm seeing about smaller forces (and therefore smaller acceleration) at the COG.
    Quote Originally Posted by pvd
    Time to stop believing the hype and start doing some science.
    29er Tire Weight Database

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