What Forces Act on a Chainring During Cycling?

[Trickystuff]

Ok, New to the forum! I'm rusty in Physics but I'm trying to find this answer...part of a personal project.

I'm trying to find the forces involved during cycling, specifically related to the Chainring (Crank), What numbers do I need to know to find this force? Let's say I have:

-Maximum power output of the rider : 1000 watts
-rpm: 100
-Crankarm length: 175mm
-Chainring radius : 75mm

Final force to the teeth of the Chainring = ?

 

[Mark44]

Ok, New to the forum! I'm rusty in Physics but I'm trying to find this answer...part of a personal project.

I'm trying to find the forces involved during cycling, specifically related to the Chainring (Crank), What numbers do I need to know to find this force? Let's say I have:

-Maximum power output of the rider : 1000 watts
-rpm: 100
-Crankarm length: 175mm
-Chainring radius : 75mm

Final force to the teeth of the Chainring = ?

The force applied to the freewheel (and hence to the wheel) will vary sinusoidally, with the maximum force occurring when the crank arm is horizontal. The minimum force occurs when the crankarm is vertical.

 

[Trickystuff]

Thanks Mark. I am trying to find threshold limits for material fatigue, so the math needs to be done and maximum load position.

 

[Merlin3189]

1000 Watt = 1000 Joule per sec
100 rpm = 100 rotations per min = 1.667 rotations per sec
So work per rotation = 1000 J per sec / 1.667 rotations per sec = 600 J per rotation

Unfortunately the rider does not apply a consistent force to the pedals. (S)he applies greater force when the crank is horizontal than when it is vertical and it probably varies in a complex way during the cycle. The next steps depend on what you think about that.

If a constant torque were applied to the pedal,
the work done per rotation = torque x 2π So 600 = T x 2π , and T = 95.49 Nm

Torque = tension in the chain x radius of chainwheel
So 95.49 Nm = tension x 0.075m and tension = 95.49 / 0.075 = 1273 N

This is an average force in the chain.

My opinion is that the force varies roughly as a sine function. If the rider is applying a vertical force on the pedals, then the torque is zero at the top and bottom of a pedal stroke, reaching maximum half way between, when the crank is horizontal.
Using this model, the average torque is about two thirds of the peak torque (average magnitude of a sine function = 2 x peak / π)
And the peak tension in the chain is 1273 x π / 2 = 2000 N (Equivalent to 200kg weight)

(Just BTW, that means the force on the pedal would be 2000N X 0.075 / 0.175 = 857 N equivalent to a weight of 87.5 kg.
A rider lighter than 87.5kg would therefore need to add to his deadweight by perhaps pulling on the handlebars.)

How the tension from the chain is distributed over the chainring is another problem. With a new chain and chainring, there may be a close fit with the tension spread over several teeth. If you lock the back wheel and apply pressure to the pedal, you could try feeling where the chain becomes free from the chainring. Even knowing how many teeth are engaged, I can't say whether the load is distributed evenly over those teeth. My guess is that the first teeth take disproportionately greater part of the load. (On my bike by the time you are half way round the chainwheel, there is no tension in the chain and I can lift it slightly from the teeth. But mine is definitely a bit worn.)

Hope my thoughts are of some use to you.

 

[zoki85]

The force applied to the freewheel (and hence to the wheel) will vary sinusoidally, with the maximum force occurring when the crank arm is horizontal. The minimum force occurs when the crankarm is vertical.

Maybe not quite: Clip-in pedals

 

[Trickystuff]

1000 Watt = 1000 Joule per sec
100 rpm = 100 rotations per min = 1.667 rotations per sec
So work per rotation = 1000 J per sec / 1.667 rotations per sec = 600 J per rotation

Unfortunately the rider does not apply a consistent force to the pedals. (S)he applies greater force when the crank is horizontal than when it is vertical and it probably varies in a complex way during the cycle. The next steps depend on what you think about that.

If a constant torque were applied to the pedal,
the work done per rotation = torque x 2π So 600 = T x 2π , and T = 95.49 Nm

Torque = tension in the chain x radius of chainwheel
So 95.49 Nm = tension x 0.075m and tension = 95.49 / 0.075 = 1273 N

This is an average force in the chain.

My opinion is that the force varies roughly as a sine function. If the rider is applying a vertical force on the pedals, then the torque is zero at the top and bottom of a pedal stroke, reaching maximum half way between, when the crank is horizontal.
Using this model, the average torque is about two thirds of the peak torque (average magnitude of a sine function = 2 x peak / π)
And the peak tension in the chain is 1273 x π / 2 = 2000 N (Equivalent to 200kg weight)

(Just BTW, that means the force on the pedal would be 2000N X 0.075 / 0.175 = 857 N equivalent to a weight of 87.5 kg.
A rider lighter than 87.5kg would therefore need to add to his deadweight by perhaps pulling on the handlebars.)

How the tension from the chain is distributed over the chainring is another problem. With a new chain and chainring, there may be a close fit with the tension spread over several teeth. If you lock the back wheel and apply pressure to the pedal, you could try feeling where the chain becomes free from the chainring. Even knowing how many teeth are engaged, I can't say whether the load is distributed evenly over those teeth. My guess is that the first teeth take disproportionately greater part of the load. (On my bike by the time you are half way round the chainwheel, there is no tension in the chain and I can lift it slightly from the teeth. But mine is definitely a bit worn.)

Hope my thoughts are of some use to you.

Wow...thanks Merlin! Let me think better through your feedback, that's awesome! I guess I can rewrite my final question: it's not the interaction to the teeth itself, my question is more on what is the final torque applied to the chairing from from perimeter to the center...now I'm thinking that I also need to know the diameter of the center fixing point on the chainring (wish is spyderless) .

 

[Mark44]

Maybe not quite: Clip-in pedals

I was thinking of those, but didn't mention them in my post. When the crankarm is vertical, the force is still at a minimum. When the crankarm is horizontal, not only do you get the downward force on the forward pedal, but you also get some upward force on the rear pedal.

 

[phyzguy]

One comment. I suggest you rethink your 1000 Watt input. Tour de France riders can put out 400-600 Watts for a short time (see this link). Most of us mere mortals have a maximum power output more like 200-400 Watts.

 

[Merlin3189]

This link http://www.cyclingpowerlab.com/CyclingPowerOutput.aspx suggests powers in the 500W area.

I tend to believe the Phyzguy reference, but some people seem to think they go higher,

" Incidentally, the very best sprinters in the Tour are going to peak at closer to 2,000 W – there are reports of Cavendish at 1,600 W and so 2,000 W for short periods is not inconceivable (a colleague of mine has tested elite BMX riders in Cape Town and measured 2,000W " in http://sportsscientists.com/2011/07/the-tour-de-france-power-2011-outputs-from-the-outside-in/

Although I'm sceptical about these values, Trickystuff does need to allow for the very peak torques if he is interested in the strength required of components.

Since we are really interested in the torque, it depends at what cadence these very high powers are reached. If these peak powers are achieved by increasing the cadence, the torque may not be higher. The nominal 100rpm is not extreme. I believe top riders go to 120 and more.

 

[phyzguy]

I agree that the peak power is really what is important here, and maybe the peak power can reach 1000 W or more. The numbers I quoted are what can be sustained for a period of minutes.

 

[A.T.]

Trickystuff does need to allow for the very peak torques if he is interested in the strength required of components.

Exactly. The maximal force on the pedal should be around the riders weight (maybe higher for highly dynamic movement). That's where you get the maximally possible torque from, not from some averaged power value. You can break a pedal at rest, thus at zero power input.

 

[Trickystuff]

Thank you all. Yes, I am interested on the peaks. Another consideration of mine is how the Chainring is configure: For example, a ring connected to a 104 BCD Spyder have a different approach on the numbers needed to find these forces. A direct mount ring (Spyderless) is one single part, and the diameter of the Slot attachment would also need to be considered...(I'm thinking) .

 

[Trickystuff]

Merlin3189, Quick last check: I've started using 1273N as my resulting number for my experiment...but after reading your message again, for some reason got me confused.

'My opinion is that the force varies roughly as a sine function. If the rider is applying a vertical force on the pedals, then the torque is zero at the top and bottom of a pedal stroke, reaching maximum half way between, when the crank is horizontal.
Using this model, the average torque is about two thirds of the peak torque (average magnitude of a sine function = 2 x peak / π)
And the peak tension in the chain is 1273 x π / 2 = 2000 N (Equivalent to 200kg weight)'...

Should I just use 1273N as my peak force concerning material strength, or 2000N? I'm a little confused on this 2000N value.

 

[Merlin3189]

2000N is the peak value. 2/3 (or really 2/π) of that is the average value 1273

See diagram

 

[Trickystuff]

My interpretation was that 1273N was found using a constant torque applied to the pedal (at 1000 Watts/100 rpm). Since I'm assigning 1000 Watts/100rpm in my question as the Maximum Power Output (not an average), this 'average' of 1273N on the chain would actually be at Peak...Thinking like if we replace the rider by a motor running at 1000 watts/100rpm.

I just realized that I didn't specified on my first question that these numbers were a Peak...not Average...I apologize for that.

 

[Merlin3189]

Yes, that's fair enough. If the peak power is 1kW, the peak torque is 1273N.

 

[Trickystuff]

Thanks you so much to all your help Merlin3189!