Author Topic: What is the force tightening a screw-on freewheel onto a hub?  (Read 1673 times)

What is the force tightening a screw-on freewheel onto a hub?
« on: 23 September, 2017, 09:13:31 pm »
I hope there are a few mathematicians who can clear my confused mind on this one.

As I see it the only factors that enter into play are the diameter of the sprocket which determines the length of the lever forcing the freewheel onto the the hub, and the force resisting the rotation of the wheel and thus blocking the hub. I don't think that size of chainring or actual gear size affects the force tightening the freewheel onto the hub. This is very poorly put but would anyone care to explain and justify the arguement (or not as the case may be).

I know this might seem trivial but I am actually in a dilemma between changing the granny ring on the Peugeot (26ring x 30 sprocket gives a 22" gear) or rebuilding the rear wheel with a hub to take a 13-34 freewheel (which with the 28t ring gives a 24" gear) History of french freewheel threads against English ones. Obviously if I can get the gearing by simply changing granny ring, with a freewheel that is easier to remove as well I will be one happy bunny.

Posted here 'cos it seems more appropriate but mods please move it if you see fit.  :)

LittleWheelsandBig

  • Whimsy Rider
Re: What is the force tightening a screw-on freewheel onto a hub?
« Reply #1 on: 23 September, 2017, 10:29:56 pm »
For the same gear (and crank length), using a smaller chainring and sprocket combination results in a higher chain tension. That increases wear rate within the chain and between the chain and sprocket but does not change the torque tightening the freewheel onto the hub. That is because the difference in chain tension is directly compensated by the change in leverage from the sprocket diameter.
Wheel meet again, don't know where, don't know when...

Karla

  • car(e) free
    • Lost Byway - around the world by bike
Re: What is the force tightening a screw-on freewheel onto a hub?
« Reply #2 on: 24 September, 2017, 12:01:08 am »
Say you were exerting the hard but not unfeasible power of 300W to travel at 20 mpg in a 100" gear on a bike with fullsize wheels.

power = force * velocity

Therefore we need to find the velocity at which the chain is being pulled forward relative to the wheel by the chain ring. 

100" x pi = 314" ≃ 8m travelled along the ground for every revolution of the pedals.  20 mph is approximately 10 m/s so our pedals will be turning at 10/8 = 1.25 Hz, i.e. a cadence of 75 rpm.

A 100" gear is a chainring/sprocket ratio of roughly 11/3, or a 44 tooth chainring to a 12 tooth sprocket.  Let's say our chainring has a radius of diameter of 20 cm.  We therefore pull chain at a rate of (20 cm * pi * 1.25) metres every second.

0.2 * 3.14 * 1.25 = Vchain ≃ 0.8m/s

Remember that  power = force * velocity, so force = power/velocity

Fchain = 300/0.8 = 375 Newtons of force on the driving part of your chain (the chain tension)

At the rear sprocket, the torque (the turning force) = force * perpendicular distance from the centre of rotation.  Since we're using a 12 sprocket, let's call that 2.5 cm.

Torque = 375 * 0.025 ≃ 9 Nm applied to your rear sprocket. 

If you use a sprocket twice as big and apply the same force to the chain, you'll be applying twice the torque to the freewheel threads.  However, if you want to go at the same speed at the same cadence with a bigger sprocket on the back, you're going to be using a bigger chainring.  This will increase the velocity of the chain pulled.  Remember that F=P/V; you've just doubled the V so halved the F.  When we come to calculate torque = FD, we've doubled the D but since the F is halved, torque stays the same. 

Re: What is the force tightening a screw-on freewheel onto a hub?
« Reply #3 on: 25 September, 2017, 04:53:36 pm »
Say you were exerting the hard but not unfeasible power of 300W to travel at 20 mpg in a 100" gear on a bike with fullsize wheels.

power = force * velocity

Therefore we need to find the velocity at which the chain is being pulled forward relative to the wheel by the chain ring. 

100" x pi = 314" ≃ 8m travelled along the ground for every revolution of the pedals.  20 mph is approximately 10 m/s so our pedals will be turning at 10/8 = 1.25 Hz, i.e. a cadence of 75 rpm.

A 100" gear is a chainring/sprocket ratio of roughly 11/3, or a 44 tooth chainring to a 12 tooth sprocket.  Let's say our chainring has a radius of diameter of 20 cm.  We therefore pull chain at a rate of (20 cm * pi * 1.25) metres every second.

0.2 * 3.14 * 1.25 = Vchain ≃ 0.8m/s

Remember that  power = force * velocity, so force = power/velocity

Fchain = 300/0.8 = 375 Newtons of force on the driving part of your chain (the chain tension)

At the rear sprocket, the torque (the turning force) = force * perpendicular distance from the centre of rotation.  Since we're using a 12 sprocket, let's call that 2.5 cm.

Torque = 375 * 0.025 ≃ 9 Nm applied to your rear sprocket. 

If you use a sprocket twice as big and apply the same force to the chain, you'll be applying twice the torque to the freewheel threads.  However, if you want to go at the same speed at the same cadence with a bigger sprocket on the back, you're going to be using a bigger chainring.  This will increase the velocity of the chain pulled.  Remember that F=P/V; you've just doubled the V so halved the F.  When we come to calculate torque = FD, we've doubled the D but since the F is halved, torque stays the same.

Yes I have thought that through, working backwards (freewheel static and force applied to wheel) to come to the conclusion that the torque must remain the same for any given resistance - so why do freewheels with bigger sprockets tighten up more for the same input to move the wheel? I am sure I am missing something somewhere (the point being that what interests me is the amount the freewheel gets tightened, not the effort to move the bicycle).

Karla

  • car(e) free
    • Lost Byway - around the world by bike
Re: What is the force tightening a screw-on freewheel onto a hub?
« Reply #4 on: 25 September, 2017, 05:50:21 pm »
If you want to apply the same power to go at half the velocity (i.e. you're riding against twice the resistance), you'll use a chainring the same size, the chain will go at the same speed and have the same tension, but the rear sprocket will be twice as large and so apply twice the torque to the freewheel thread.

Changing down into a more 'torquey' gear to ride in the same conditions isn't going to change anything, because you'll either put out less power or else you'll spin faster, each of which put the chain under less tension.  A low gear will however allow you to ride in higher resistance conditions by multiplying the amount of torque applied by your legs.  The tradeoff is that you go slower. 

JennyB

  • Old enough to know better
Re: What is the force tightening a screw-on freewheel onto a hub?
« Reply #5 on: 25 September, 2017, 07:26:54 pm »
Therefore, for the same rider, course, an gear (in inches) the freewheel on a Brompton doesn't get screwed on as tightly?
Jennifer - Walker of hills

LittleWheelsandBig

  • Whimsy Rider
Re: What is the force tightening a screw-on freewheel onto a hub?
« Reply #6 on: 26 September, 2017, 06:01:47 am »
Correct, small wheels reduce the torque 'experienced' by the components in/on the rear wheel (when compared to big wheels).
Wheel meet again, don't know where, don't know when...