Force in a chain?

That was a perfect explanation, JamesM 8)

It's fascinating how much can be said about a thing with a math behind as simple as gtRTSdh has already shown, so may I add some more :lol: Btw. reading those mechanical explanations/conversations always remembers me a bit of those antique competitions of how to construct a certain perpetuum mobile...

The solution to the chainforce question has already been mentioned many times, it depends only on the size of the chainwheel. However, I might have found suburban's point to reason differently, i.e. that the chainforce depends on the gear ratio. So I thought I made a little draft to add to the diversity, and I called the forces

F1...riders force on pedal,
F2...chain force and
F3...accelerating force at the rear wheel

and it turns out, that you might see the point of misunderstanding between "chainring size" and "gear ratio", if you somewhat equalize / interchange F2 and F3 in your thoughts.

The torques are the masters of the game, so have fun:
 

Attachments

  • Chainforce_question_sm.webp
    Chainforce_question_sm.webp
    85 KB · Views: 376
gtRTSdh":26nzapzz said:
The maths is quite simple;

Looking at the situation when maimum torque is applied and there is sufficient traction. With drive side crank being loaded with the force of the riders weight/strength at the horizontal position (maximum leverage), ignoring the small (vertical) difference in angle of the chain when on size different sprockets at the rear.

Chain force = riders weight x [ crank length (mm) / chainring radius (mm) ]

100Kg rider 24T granny ring, 44T big ring.

So 100Kg x [175mm/45mm ( radius of 24T chainring)] = 388.8 Kgs
So 100Kg x [175mm/90mm ( radius of 44T chainring)] = 194.4 Kgs

Really? Are you sure about that??

http://www.scribd.com/doc/62497413/Powe ... -and-Chain

Pages 31-34... :wink:

(Loving this thread BTW - threads like this that make life interesting... :lol: )
 
if I may answer: yes the math is really that simple (it's the same formula I showed in my drawing - beside using kgs as unit for force).
You can blow up anything, which happens btw. in many papers/books so that people don't understand what the author is talking about, just as you can see in your link. Here the author uses tons of "safety factors" instead of a proper error analysis, which incredibly clutters the math, so that you hardly "see" what the formula is really about. He also discusses time (which is irrelevant for us), so he introduces to nearly every variable the according "velocity" as new variable instead of "per second". No wonder, the book just denies to deal correct in any way with units. These books are also example for introducing new variables of the form

new variable = 1 / known variable --> makes one more variable

I love those! That's what you do if you have too much time, or if you consider your audience to have too much time, or if you are just a smart guy whom nobody understands... :wink:
 
02gf74":23eah0rb said:
Question is about the the force in a chain on a bike - which one would have bigger force going through it:

1. driving gear 39 T, driven gear 23 T (e.g. road bike)
or
2. driving gear 22 T, driven gear 32 T (e.g. mountain bike)

The motor unit (i.e. the cyclist) is generating the same amount of torque.

I also assume the greatest force is in the lowest gear - is that correct?

I was answering the original question. I used Kg's so it's simple to understand. Force varies between each rider, their, weight, strength and effort, point of pedal stroke at a given time. All but the riders weight are nearly impossible to ascertain (at home) and of course force varies constantly throughout the 360 pedal stroke. It would also be sensible to add the additional upward force of the left leg on the rear pedal, but that's too many unknown variables to actually get an answer with the available information.

Acceleration is also something that is very difficult to measure (at home) on a bike. So apply a little common sense, consider the situation stationary at the maximum leverage point using the known variables, compare two chainring radii, and you get to a simple ratio of levers which answers the OP's question.

I refer to my point, that it's impossible (for a human leg) to acheive maximum Force in a short instant as with an easy gear,

'There is also the point to consider that in the easiest gear, because the gear is so easy to pedal the rider does not have time to achieve full muscle strain, which is why it you can pedal with more force at lower cadences.'

But it is possible to stand up and heave on 22front - 11rear from stationary so that really only applies to very easy gears.
 
We_are_Stevo":1psy33uk said:
gtRTSdh":1psy33uk said:
The maths is quite simple;

Looking at the situation when maimum torque is applied and there is sufficient traction. With drive side crank being loaded with the force of the riders weight/strength at the horizontal position (maximum leverage), ignoring the small (vertical) difference in angle of the chain when on size different sprockets at the rear.

Chain force = riders weight x [ crank length (mm) / chainring radius (mm) ]

100Kg rider 24T granny ring, 44T big ring.

So 100Kg x [175mm/45mm ( radius of 24T chainring)] = 388.8 Kgs
So 100Kg x [175mm/90mm ( radius of 44T chainring)] = 194.4 Kgs

Really? Are you sure about that??

http://www.scribd.com/doc/62497413/Powe ... -and-Chain

Pages 31-34... :wink:

(Loving this thread BTW - threads like this that make life interesting... :lol: )

We don't have enough information to use those formula's, which assume an electric motor driving, which have a constant torque output which makes life very easy when doing calculations.

A riders torque output oscillates from practically zero at 12-6 O'clock position to maximum torque at 3-9O'clock position. This makes the fomulas unsuitable for our requirement. Of course you can take averages of rider power and torque, but the OP asked about maximum force.
 
a) Why are granny rings - often - made of steel?
b) I have broken chains going from middle to granny (going up hill I might add). See (d) for solution.
c) What would Sir Chris Hoy say?
d) Remove granny. Strips horrible heavy steel from the bike and makes legs like the bloke mentioned in (c)

Apologies I can't contribute to something more theoretical, but so far those two pictures of lots of maths look very convincing.
 
Woz":2p6f6m56 said:
a) Why are granny rings - often - made of steel?
b) I have broken chains going from middle to granny (going up hill I might add). See (d) for solution.
c) What would Sir Chris Hoy say?
d) Remove granny. Strips horrible heavy steel from the bike and makes legs like the bloke mentioned in (c)

Apologies I can't contribute to something more theoretical, but so far those two pictures of lots of maths look very convincing.

I can answer the granny ring bit.

It make the rings last longer.
-There is increased pressure on each tooth/chain interface as there are less of them in contact with the chain at anyone time, so grind down quicker.
-You could also argue they revolve more often if you wanted.
-They are cheaper to make so come on lower groupsets ;)
 
UNCLEJACK":3shtobnk said:
So what happens when you start "BACK PEDALLING?" :lol:

I didn't, I stuck to my guns the whole time. For two different sized gears sets with the same ratio the chain tension will be higher on the smaller set, all else being equal :wink: :lol:
 

Latest posts

Back
Top