# Thread: why do bigger disc's equal better stopping??

1. ## why do bigger disc's equal better stopping??

Sorry for the stupid question but I am realitively new to the disc brake world.

Why, all things equal (best I can tell is the same model with 160mm vs 185mm still uses the same pads, caliper etc) does a bigger disc diameter equal better stopping. The pads still only bare on the outer1/4 of the disc.

Is it that there is more baring surface due to the larger diameter or better heat discipation, both (neither lol).

Thinking about putting a 185mm (up from 160mm) upfront for next year (with the correct bracket ofcourse) so I want to make sure I know why it works better, lol

J-

2. The same pad and caliper grip the same amount of disk with the same pressure....

Thing is the bigger disk can dissipate heat better...

and it has a bigger lever arm to work with.

The heat dissipation is the big thing.

3. The unscientific answer is it takes less energy to stop a spinning wheel closer to the outside of its diameter. A larger disc effectively moves the braking area closer to the outside of the wheel.

Someone will chime in on the science....

Cheers!

4. Originally Posted by cmdrpiffle
The unscientific answer is it takes less energy to stop a spinning wheel closer to the outside of its diameter. A larger disc effectively moves the braking area closer to the outside of the wheel.

Someone will chime in on the science....

Cheers!
The energy required to stop the bike remains the same regardless of the disk diameter.

5. Originally Posted by jeffscott
The energy required to stop the bike remains the same regardless of the disk diameter.
Yes but you have a better mechanical advantage with a bigger rotor. It's not solely a matter of heat although it also plays a role. If we take heat out of the equation, the bigger rotor will still have more power.

6. One word: torque

7. Originally Posted by PissedOffCil
Yes but you have a better mechanical advantage with a bigger rotor.Yes a longer (bigger) lever arm It's not solely a matter of heat although it also plays a role. If we take heat out of the equation, the bigger rotor will still have more power.
You are mixing your terms.....Power refers to energy per unit time. The brake converts kinetic power, to heat...and it must dissipate or store this heat. The bigger rotor will be better for both dissipation and storage of heat. So heat dissipation and storage provide more braking power.

The lever arm provides greater braking torque because of the longer lever arm.

End result bigger rotor equals more braking torque and more braking power.

For me a 160 mm rotor provided ample one finger braking torque...(I could do a really good stopee)....but on long descents I would smoke the brake and lose braking torque due to overheating.

So for me the bigger rotor 203mm improved the braking power by increasing heat dissipation/storage.

8. One word: Leverage (which basically means torque)

9. Originally Posted by sandmangts
One word: Leverage (which basically means torque)
+1 think of it as a lever on a fulcrum and you are trying to lift a large weight if you have 10ft of lever on your side of the fulcrum it will be easier to lift the weight then if you have 5ft on your side

Did i make any sense?

so if you have a larger rotor you then have a larger lever so to speak which makes it that much easier to stop the bike.

10. More surface area to compress upon...

"bigger is better"....oh, baby!

11. cool thanks guys all makes sense and was kinda what I was thinking.

Second quick question. To go to a bigger front rotor all I need is the rotor itself and the appropriate caliper adaptor?

Same cailper/pad would work, correct? I have BB5's, if that matters.

Thanks
J-

12. Yes, make sure to get the proper FRONT adaptor.

Yes, make sure to get the proper FRONT adaptor.
copy thanks all for ur input.

J-

14. Check if your front fork will support the bigger rotor. I wanted to upgrade but RockShox said that my fork wouldn't support the bigger rotor. I assume the fork didn't have enough support for the extra stopping power or something.

Check if your front fork will support the bigger rotor. I wanted to upgrade but RockShox said that my fork wouldn't support the bigger rotor. I assume the fork didn't have enough support for the extra stopping power or something.
copy i'll def check that before I lay out the cash.

thanks

J-

16. It's torque and surface area. The torque has been explained. But on the larger diameter, more rotor passes through the caliper with one rotation of the wheel. More circumference = more brake length for the same distance traveled. A 160mm has 502.654mm of braking length and a 185mm has 581.946mm, that's an extra 3" of brake per revolution.
Btw, circumference = pi x diameter.

Oh yeah, and better heat dissipation.

17. Originally Posted by jeffscott
You are mixing your terms.....Power refers to energy per unit time. The brake converts kinetic power, to heat...and it must dissipate or store this heat. The bigger rotor will be better for both dissipation and storage of heat. So heat dissipation and storage provide more braking power.
I'm not mixing terms at all. Mechanical advantage is the same as leverage and this is what gives you better stopping power with a bigger rotor.

From Sheldon :
"'Mechanical advantage', or 'leverage' is the ratio between how much you get out of a linkage and how much you put in."
"Installing smaller wheels [... reduces] the mechanical advantage accordingly. For instance, substituting 622 mm (700C) wheels on a bike built for 630 mm (27 inch) wheels will degrade the braking."

The same applies to rotors.
Ref : The Geometry of Cantilever Brakes

18. Originally Posted by PissedOffCil
I'm not mixing terms at all. Mechanical advantage is the same as leverage and this is what gives you better stopping power with a bigger rotor.

From Sheldon :
"'Mechanical advantage', or 'leverage' is the ratio between how much you get out of a linkage and how much you put in."
"Installing smaller wheels [... reduces] the mechanical advantage accordingly. For instance, substituting 622 mm (700C) wheels on a bike built for 630 mm (27 inch) wheels will degrade the braking."

The same applies to rotors.
Ref : The Geometry of Cantilever Brakes
He's trying to point out that you are using the word "power" in the wrong context. Notice how neither of the statements you just listed reference "power"?

19. Originally Posted by kan3
He's trying to point out that you are using the word "power" in the wrong context. Notice how neither of the statements you just listed reference "power"?
No his edited quote mentions that a longer lever gives better mechanical advantage, which it does but a bigger rotor also gives a better mechanical advantage.

Yes I used "power" as in "stopping power". Given the context I think it goes without saying and without getting into semantics.

20. nice ^ ! way to sum it all up.

21. Originally Posted by PissedOffCil
No his edited quote mentions that a longer leverlever arm as in the radius between the brake pad and the axle gives better mechanical advantage, which it does but a bigger rotor also gives a better mechanical advantage.

Yes I used "power" as in "stopping power". Given the context I think it goes without saying and without getting into semantics.
Nope still got it wrong but what the hey....you at least can understand yourself.

22. Have you ever extended the handle of a tire iron to make it easier remove lug nuts?

I believe a bigger rotor works on the same principle to provide more stopping power.

23. Originally Posted by JanBoothius
Have you ever extended the handle of a tire iron to make it easier remove lug nuts?

I believe a bigger rotor works on the same principle to provide more stopping power.
Yup the extended handle, gives you a longer lever arm....hence more torque....not more power.

Anyway we call that a snipe around here.

24. Originally Posted by jeffscott
Nope still got it wrong but what the hey....you at least can understand yourself.

What exactly are you talking about? Are you saying that a bigger rotor doesn't have a bigger mechanical advantage over a smaller rotor?

25. Originally Posted by jeffscott
Yup the extended handle, gives you a longer lever arm....hence more torque....not more power.

Anyway we call that a snipe around here.
You're really just playing with words. When people talk about stopping power, they mean the ability to stop in a shorter amount of time/distance from a given speed. You can go around correcting people with your torque vs. power all you want, it really makes you look like a dork... Everybody here understood what we are talking about, we couldn't care less about your semantics and uptight comments.

26. Originally Posted by PissedOffCil

What exactly are you talking about? Are you saying that a bigger rotor doesn't have a bigger mechanical advantage over a smaller rotor?
No not at all that is correct...

The bigger rotor locates the caliper and pad further away from the axle....this results in a longer "lever arm" for the same force to act upon....

Hence more braking torque (notice the words "lever arm", that is a precise engineering term to describe the length of the radius upon which a force acts to develop a torque....not the brake lever).

Second Brake Torque is not Brake Power.

A brake must dissipate the Kinetic Power into Heat without overheating and fading.

That is governed by heat storage, and heat dissipation both of which are also higher with a bigger rotor.

27. Originally Posted by PissedOffCil
You're really just playing with words. When people talk about stopping power, they mean the ability to stop in a shorter amount of time/distance from a given speed. You can go around correcting people with your torque vs. power all you want, it really makes you look like a dork... Everybody here understood what we are talking about, we couldn't care less about your semantics and uptight commentsClearly some do, also clearly you don't.
Not at all the point is that brake torque and brake power are two very different things is quite important...

For example with Shimano's ICE technology you can get a smaller, lighter brake (smaller rotor diameter) with equivalent power (because they have installed a heat sink on the pads)to a brake with a larger rotor, but less brake torque (as long as the brake levers etc are the same).

Precise language often helps clarify ones understanding of issues

28. More metal passed through the friction material ( pads ) for a given wheel rotation, which ='s more friction, simple.

Bigger rotors also heat up slower cause they generally have more material to heat and disipate the heat faster due to the higher surface area.

29. Math's wise the faster your going the more energy you need to be stopped.

Kinetic Energy = weight * ( Speed ^ 2 )

Thankfully as you get faster your friction increases.

Friction = Rotor Speed ^ 2

So you have a equal lever force requirement at 30mph as you do at 15mph pretty much ommiting some small variables. Despite at 30mph having 4x's the kinetic energy.

30. Originally Posted by Turveyd
More metal passed through the friction material ( pads ) for a given wheel rotation
^ This is what it all boils down to. Given the same calipers (implying equivalent "squeezing" force) and the same rotor and pad materials (implying an equivalent coefficient of friction), the only difference is that for one wheel rotation the caliper will be applying friction to the larger rotor across a longer distance. For any given speed, the caliper needs to apply it's frictional force through the same distance (the arc length of the rotor in this case) regardless of the rotor size; the difference between the two cases is that for a given arc length, the smaller rotor needs to go through more full rotations than the larger rotor, which means that the wheel will needs to go through more rotations and you will travel farther before coming to a stop.

31. Originally Posted by TheNightman
^ This is what it all boils down to. Given the same calipers (implying equivalent "squeezing" force) and the same rotor and pad materials (implying an equivalent coefficient of friction), the only difference is that for one wheel rotation the caliper will be applying friction to the larger rotor across a longer distance. For any given speed, the caliper needs to apply it's frictional force through the same distance (the arc length of the rotor in this case) regardless of the rotor size; the difference between the two cases is that for a given arc length, the smaller rotor needs to go through more full rotations than the larger rotor, which means that the wheel will needs to go through more rotations and you will travel farther before coming to a stop.
Yep, most people think it's leverage how far the caliper is out, although leverage is a part, the bigger the lever the more pressure you can apply to the pads, they can't call it a brake lever for nothing

29" Wheels, need 10% bigger rotors for the same power as you travel 10% further for the same amount of friction, 29ers with 160's rotors is the same as 26" with 144mm rotors power wise, not heat wise though still 160's.

32. Energy is the ability to do Work. Work is Force x Distance x the Cosine of the angle. The pads always make contact at 90 degrees to the hub, and Cos 90 is 1.

So W = FD

The only way for a brake to do more work on the rotational energy (momentum or mass times velocity) of a spinning wheel is to either increase the Force (4 piston brakes, better design, etc) or increase the distance of the lever arm. A lighter wheel has less momentum since it has less mass but that's off topic...

What's next? Rim brakes?

33. Originally Posted by SJDude
Energy is the ability to do Work. Work is Force x Distance x the Cosine of the angle. The pads always make contact at 90 degrees to the hub, and Cos 90 is 1.

So W = FD

The only way for a brake to do more work on the rotational energy (momentum or mass times velocity) of a spinning wheel is to either increase the Force (4 piston brakes, better design, etc) or increase the distance of the lever arm. A lighter wheel has less momentum since it has less mass but that's off topic...

What's next? Rim brakes?
4 pistons on there own do not equal more power, they may allow you to have a longer pad which isn't as deep which sits further out on your rotor and gives you a fraction more power also the 1 side can generally move before the other to get rid of snatch at first then you can run more pressure to the piston.

say 400grams off both rims / tyres compared to a 200lb person and bike which is 90,800grams allowing for the rotation is likely 0.6% factor in aerodynamics as your speed increases and this drops considerably.

Brake Force = ( ( [1/2 Pad Radius] * 2 ) * Pi ) * [Friction] ) * ([Rotational Speed]^2)

[Friction] being a variable 0 for not in use and up ofcourse,

Hence, narrower longer pad gains you a few mm's in effective rotor size.

If you used a Planetary gear system on your hub to disk you could double the rotor speed and get double the power, but at a weight cost.

34. Now let's factor in pad material...

35. Originally Posted by Turveyd
.........................

If you used a Planetary gear system on your hub to disk you could double the rotor speed and get double the power, but at a weight cost.
That is actually ingenious. Overly complicated but should work.

36. ^^^^
Hammerschmit idea + disc brakes = downhills newest money pit

37. I was curious how this question rcvd so many responses. Now I know and I want my 2 minutes back. ; )

38. [DEleted]

39. not entirely sure, but i believe the bigger the rotor, the more rotor contact with the pads, allowing for better stoping power.

Also, the ability of the larger disks, to have more time to cool down, inbetween points.. So like lets say a 180mm disk takes 1 sec to do a full rotation (A to B).. So 260mm disk takes about 3sec to do a full rotation (A to B)..

That is my belief.

40. It's not contact with Pads area although bigger helps, as your same force through the lever is just spread over a wider area.

And Pad Material changed the Friction value, increasing or decreasing.

And yes, having your rotor spinning at 100mph when your doing 20mph would look cool!!

41. Originally Posted by Turveyd
And yes, having your rotor spinning at 100mph when your doing 20mph would look cool!!
Right until you fall off and get a finger caught in it.

Or until the caliper explodes...

Although, having a gearbox where the casing rotates with the output drive would be awesome!

42. Originally Posted by Fix the Spade
Right until you fall off and get a finger caught in it.

Or until the caliper explodes...

Although, having a gearbox where the casing rotates with the output drive would be awesome!
Well always downsides, but proves nicely it's not the stupid leverage most people think it is.

Wouldn't generate any more heat maybe cool better as it's spinning so fast, heat is just kinetic energy of your movement being transfered to heat, so speed = same heat generated.

43. Like I said, it's the amount of rotor that travels between the pads per rotation that's the rotor difference (brake length/circumference). That's why the planetary would work. Not sure why everyone had to overthink it. Sure leverage plays a part, but not as much as the extra 3" of rotor passing through the caliper from 160 to 180.

44. Originally Posted by curtboroff
Like I said, it's the amount of rotor that travels between the pads per rotation that's the rotor difference (brake length/circumference). That's why the planetary would work. Not sure why everyone had to overthink it. Sure leverage plays a part, but not as much as the extra 3" of rotor passing through the caliper from 160 to 180.
I'm with you on that, the only place where leverage comes into it on a disk brake system is you guessed it the Brake lever, thats why it's called a lever, the leverage there lets you compress the pads via the fluid.

You can't science wise use Leverage based on distance from QR, although the calc works out the same it makes no allowance for Friction and assumed brake always on or ability to increase the power relative to the riders speed.

People are moron's and repeat the first thing they hear with no ability to change or rethink it most of the time. Even when the subject is obviously way above there tiny little brains.

45. Sorry it's a arguement I had on another forum with everyone stick and so called mates, still sniggggers ( can't say the N word LOL ) and say leverage like they are correct even today 1 had a pop today and said unable to admit when I'm wrong, which is kinda ironic as he's wrong.

People!!

46. Originally Posted by Turveyd
I'm with you on that, the only place where leverage comes into it on a disk brake system is you guessed it the Brake lever, thats why it's called a lever, the leverage there lets you compress the pads via the fluid.The lever arm referred to is the mechanical advantage of the system

You can't science wise use Leverage based on distance from QR, dam right you can although the calc works out the same it makes no allowance for FrictionWhat units and assumed brake always on or ability to increase the power relative to the riders speed.

People are moron's and repeat the first thing they hear with no ability to change or rethink it most of the time. Even when the subject is obviously way above there tiny little brains.
Same difference boys...think it through.....Oh and BTW brakes still overheat....

47. Holy Cow, havent checked in this thread in a while, suffice to say my question has been anwered, LOL

Got an 180mm adaptor from North Shore Billet on the way and picking up a 180mm avid HSX two piece rotor from Tree Fort tomorrow.

J-

48. Originally Posted by jeffscott
Same difference boys...think it through.....Oh and BTW brakes still overheat....
There calc, the caliper would have to move in and out to vary the power LOL

And 100mm caliper position with a bolt through or 200mm with a bolt through both has exactly the same effect on the wheel, it won't move, Leverage can only count when the disk is locked solid within the caliper.

Thats only then useable to work out the strength needed from QR to Caliper.

49. Originally Posted by Turveyd
There calc, the caliper would have to move in and out to vary the power LOL

And 100mm caliper position with a bolt through or 200mm with a bolt through both has exactly the same effect on the wheel, it won't move, Leverage can only count when the disk is locked solid within the caliper.

Thats only then useable to work out the strength needed from QR to Caliper.
Do you actually know what you're talking about? You seen to be very forthright and you seem to be suppressing others' contributions.

On this basis I won't join in the debate any further than to say I think you are incorrect on a number of levels and your equations are pretty much all bogus.

Leverage? I'd say think again because you're way off base. Thing is you are so convinced of your correctness I can't see room for a reasoned debate.

Just saying.

Sent from my GT-I9100 using Tapatalk

50. Saying leverage has nothing to do in the equation is stupid.

51. Starting from the beginning:
breaking is applying force by friction.
This force is transferred from the disc to the ground through torque.
The torque is calculated as: torque = (caliper traction force) * (distance from wheel center)

So for a given caliper, the bigger the distance of the caliper from the wheel center, the bigger the torque applied!

Now, as other users mentioned above, there are other factors that come in the equation "breaking" such as heat dissipation, clamping caliper force, area of contact, materials etc. Those factors directly influence the force applied to the disc and can equally affect the breaking power (torque).

Heat Dissipation
the more heat you can send out of the caliper-disc group, the more constant will be the breaking.
Clamping Force
it has to do with the force applied from the caliper itself from the hydraulic system. The force from our fingers is transferred from the hydraulic pump on the lever, through the pipe to the caliper piston. The size of the hydraulic pump, the lever geometry, the pipe stiffness, the hydraulic liquid, the piston diameter and the stiffness of the caliper have their role on the force transfer from finger to caliper.
Contact Area
The contact area transferrs the force applied from the caliper to pressure! For a given caliper force, the bigger the contact area, the less the pressure applied and the bigger the heat transfer gets.

Now, it gets very complicated not only for us but for the designers and manufacturers of breaks to understand all these parameters and fit them to a bike with the less weight possibble having the bigger breaking torque applied.

Bottom line
If one chooses a break type (caliper, lever) that fit ones needs, for a bigger force, will choose bigger disc and for lesser force smaller disc.

I hope this description sums up all the above.

52. I hope we can move on, this thread should have ended at the 2nd post.

53. The primary function of a braking system is FRICTION not leverage, without friction there is no leverage, with a change of friction the leverage changes so why bother with the out dated leverage static calculation.

140mm or 203mm if you can apply enough force to lock the disk the effect on the front is both equal you go over the bars, so more 'leverage' same result, proves how pointless doing is as a leverage calculation is.

54. Originally Posted by gfourlias
Starting from the beginning:
breaking is applying force by friction.
This force is transferred from the disc to the ground through torque.
The torque is calculated as: torque = (caliper traction force) * (distance from wheel center)

So for a given caliper, the bigger the distance of the caliper from the wheel center, the bigger the torque applied!

Now, as other users mentioned above, there are other factors that come in the equation "breaking" such as heat dissipation, clamping caliper force, area of contact, materials etc. Those factors directly influence the force applied to the disc and can equally affect the breaking power (torque).

Heat Dissipation
the more heat you can send out of the caliper-disc group, the more constant will be the breaking.
Clamping Force
it has to do with the force applied from the caliper itself from the hydraulic system. The force from our fingers is transferred from the hydraulic pump on the lever, through the pipe to the caliper piston. The size of the hydraulic pump, the lever geometry, the pipe stiffness, the hydraulic liquid, the piston diameter and the stiffness of the caliper have their role on the force transfer from finger to caliper.
Contact Area
The contact area transferrs the force applied from the caliper to pressure! For a given caliper force, the bigger the contact area, the less the pressure applied and the bigger the heat transfer gets.

Now, it gets very complicated not only for us but for the designers and manufacturers of breaks to understand all these parameters and fit them to a bike with the less weight possibble having the bigger breaking torque applied.

Bottom line
If one chooses a break type (caliper, lever) that fit ones needs, for a bigger force, will choose bigger disc and for lesser force smaller disc.

I hope this description sums up all the above.
You can try and look as smart as you want but you look like a dolt spelling brake "break"

55. Originally Posted by Ilikebmx999
You can try and look as smart as you want but you look like a dolt spelling brake "break"
We completely ignore him due to a spelling eror then, just cause it doesn't agree with your thoughts.

Forums and People!!

56. Originally Posted by Turveyd
The primary function of a braking system is FRICTION not leverage, without friction there is no leverage, with a change of friction the leverage changes so why bother with the out dated leverage static calculation.

140mm or 203mm if you can apply enough force to lock the disk the effect on the front is both equal you go over the bars, so more 'leverage' same result, proves how pointless doing is as a leverage calculation is.
Doh Given the same "friction" (more precisely frictional force) ie same squeeze of the pad on the rotor.

The bigger rotor applies more braking torque....

By 203mm/140mm or 1.45 more braking torque...

quite simple actually.

57. Originally Posted by Turveyd
The primary function of a braking system is FRICTION not leverage, without friction there is no leverage, with a change of friction the leverage changes so why bother with the out dated leverage static calculation.

140mm or 203mm if you can apply enough force to lock the disk the effect on the front is both equal you go over the bars, so more 'leverage' same result, proves how pointless doing is as a leverage calculation is.
Except for a minor detail: you don't need the same force at the lever to lock different size rotors...

Now lets move on shall we?

58. Originally Posted by jeffscott
Doh Given the same "friction" (more precisely frictional force) ie same squeeze of the pad on the rotor.

The bigger rotor applies more braking torque....

By 203mm/140mm or 1.45 more braking torque...

quite simple actually.
And the increased metal from the rotor moving through the pads increasing the friction created is nothing to do with it then.

Increased friction = more torque, primary is still FRICTION!!

59. Originally Posted by Turveyd
And the increased metal from the rotor moving through the pads increasing the friction created is nothing to do with it then.

Increased friction = more torque, primary is still FRICTION!!
Try to say it properly and you will have the answer....

I have already done the math and there it is:

203/140= 1.45 times more torque

203/160 = 1.27 times more torque

203/180 = 1.13 times more torque

180/140 = 1.28 times more torque

180/160 = 1.12 times more torque

160/140 = 1.14 times more torque...

Heat storage will follow the same rule pretty closely....

Heat Transfer is a little more complicated but again use the same rule....

60. Originally Posted by jeffscott
Try to say it properly and you will have the answer....

I have already done the math and there it is:

203/140= 1.45 times more torque

203/160 = 1.27 times more torque

203/180 = 1.13 times more torque

180/140 = 1.28 times more torque

180/160 = 1.12 times more torque

160/140 = 1.14 times more torque...

Heat storage will follow the same rule pretty closely....

Heat Transfer is a little more complicated but again use the same rule....
I think I know very well what the maths work out at, it's the application which is in error, your method has zero ability to increase the power as your speed increases.

Yours is a static calculation mine is dynamic, mine increase the power relative to the speed of the bike, remember the bike as it goes faster how more kinetic energy,

Remember E = MS^2 so at 20mph you require 4 times the braking force as 10mph.

Mine is the proper calculation!!

61. Originally Posted by Turveyd
And the increased metal from the rotor moving through the pads increasing the friction created is nothing to do with it then.

Increased friction = more torque, primary is still FRICTION!!
What units [FRICTION!!] ?

Talking about primary or not is a pointless. Think in terms of frames of reference and you don't need to be a big bully claiming your view is more important than anyone else's.

What units [FRICTION!!] ? In your equation:
Originally Posted by Turveyd
Brake Force = ( ( [1/2 Pad Radius] * 2 ) * Pi ) * [Friction] ) * ([Rotational Speed]^2)
...it turns out to be MASS (S.I. unit kg).

In what text book did you ever see friction described as a mass? Basically you've got the basics completely basically wrong. So stop making something really simple complicated.

Braking converts kinetic energy to heat. More effective braking converts more kinetic energy per second. Therefore the braking POWER needs to be high (POWER=ENERGY/SECOND).

POWER=TORQUEx ANG.VELOCITY

i.e.

In the question asked by the O.P. we can see that ANG.VELOCITY and FRICTION.FORCE remain the same. The only thing that changes is the ROTOR.RADIUS.

Now it also happens that ROTOR.RADIUSxANG.VELOCITY = the linear velocity of the rotor past the pads. So by changing the frame of reference you get an equally valid equation much closer to the one you have been going on about:

POWER=FRICTION.FORCExLINEAR.VELOCITY ...only this is much less useful for telling you what change you get from changing the rotor radius

You have been promoting an argument about primacy between two variations on the same equation (and coming up with spurious stuff besides).

I'm with Pissedoffcil. This thread should have stopped after post 2.

62. Originally Posted by Turveyd
I think I know very well what the maths work out at, it's the application which is in error, your method has zero ability to increase the power as your speed increases.

Yours is a static calculation mine is dynamic, mine increase the power relative to the speed of the bike, remember the bike as it goes faster how more kinetic energy,

Remember E = MS^2 so at 20mph you require 4 times the braking force as 10mph.Energy is not force

Mine is the proper calculation!!
Braking torque is what I calculated....Power is proportional to RPM X torque so there you go.....

Oh and BTW Energy is not the same as force so you need to work on you engineering skills

63. Originally Posted by petercarm
I'm with Pissedoffcil. This thread should have stopped after post 2.
In fact it should have been post 3 since post 2 only talked about heat, missing a big deal of the physics.

64. Hey I was just thinking about this whole leverage and braking thing. Don't you stop faster when putting the most pressure on the brakes WITHOUT locking the wheel?
If this is true, then having a larger diameter rotor allow you to squeeze your brakes harder without locking the wheel, therefore creating more heat from friction since you'd be squeezing the brakes harder. I thought of this by thinking of a bike upside down, if you locked a 160mm brake and tried moving the wheel by gripping the tire, then you'd have a lot of leverage over the rotor, but a bigger rotor would reduce your leverage, requiring more braking force to lock it up.

...I hope this makes sense, I'm just a 17 year old who loves to bike LOL

65. Originally Posted by justinbuzar
Hey I was just thinking about this whole leverage and braking thing. Don't you stop faster when putting the most pressure on the brakes WITHOUT locking the wheel?
If this is true, then having a larger diameter rotor allow you to squeeze your brakes harder without locking the wheel, therefore creating more heat from friction since you'd be squeezing the brakes harder. I thought of this by thinking of a bike upside down, if you locked a 160mm brake and tried moving the wheel by gripping the tire, then you'd have a lot of leverage over the rotor, but a bigger rotor would reduce your leverage, requiring more braking force to lock it up.

...I hope this makes sense, I'm just a 17 year old who loves to bike LOL
100% the opposite way, advantage of a bigger rotor is you haven't got to squeeze as hard then you've got more hand control and less likely to lock it up, but if required you can lock it up easier, if you get the desire to do a superman over your bars that is

66. Originally Posted by Turveyd
100% the opposite way, advantage of a bigger rotor is you haven't got to squeeze as hard then you've got more hand control and less likely to lock it up, but if required you can lock it up easier, if you get the desire to do a superman over your bars that is
Ohhh ok thanks! I knew I was on to something haha

67. Originally Posted by justinbuzar
Hey I was just thinking about this whole leverage and braking thing. Don't you stop faster when putting the most pressure on the brakes WITHOUT locking the wheel?
If this is true, then having a larger diameter rotor allow you to squeeze your brakes harder without locking the wheel, therefore creating more heat from friction since you'd be squeezing the brakes harder. I thought of this by thinking of a bike upside down, if you locked a 160mm brake and tried moving the wheel by gripping the tire, then you'd have a lot of leverage over the rotor, but a bigger rotor would reduce your leverage, requiring more braking force to lock it up.

...I hope this makes sense, I'm just a 17 year old who loves to bike LOL
cool cool, i am also 17, who loves to bike.

Anyways, i dont think the pressure is really related to how well you can stop with a larger dia. disc. Because no matter what, i believe the lever preasure is more of a linear friction onto the disc, no matter what size the Dia. of the disc is..

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