Gear Ratio drop in Weight (or pedal force) Savings?
I have been a avid mountain biker for many years but not until now did I question the Gear Ratios. I have read hundreds of Gear Ratio infos on the web in the past 2 weeks and could not find my answer, I wish you can help. Say, I (175lb and bike is 25lb) am riding 44 teeth front and 11 teeth at rear on perfectly flat road; as I shift rear gear to 13 teeth, I immediately feel lighter (less pedal force) while slow down a bit.
Question: How much lighter in lb? or to be more exact, How much lighter in lb when Gear Ratio drop 1 unit?
All bikers and mathematicians/Physicians please kindly advise. A formula would really help me! Thanks!!
Sorry no details, typed & polished a bunch o fluff, then lost it when tried to send over a dead connection. (EDIT - I came back & added some detail )
Physics definitions. Work statement:
Work = force * distance or, for angular motion, Work = torque * angular displacement. (gosh I hope that's right )
Equivalent power statement (power output equals time rate of doing work):
Power = force * velocity or, for angular motion, Power = torque * angular velocity.
If going the same speed, then to a good approximation, the bike is using the same power, same work per time interval. From the above, not too hard to show that changes in average force are proportional to changes in the gear ratio. (Sorry, that's the part that got lost, and I aina doin it again )
For shifting to the 13t cog from the 11t, force is 11/13. Suppose shifting the front from 40 to 52, force is 52/40.
This ignores mechanical friction changes. These are probably small so can be ignored for this discussion.
Also neglects biomechanical factors - accelerating & decelerating your thigh, calf & foot - consensus seems to be that 80 rpm is the sweet spot for most people - anyways this is beyond the scope of the question.
Can't know 'exact number of pounds' without a whole lot more info, some of it difficult to measure. Unless you have one of those BBs that measures it directly. I spose could make a stab at it -- need to know the crank arm length, the average number of watts it takes to propel an average (mtn/road/take yer pick) bike at (pick your speed), could take the diagram I saw years ago of average force profile for the power stroke and do some integral calculus on it... to estimate force at a specific angular position ... but life is too short...
Last edited by duffer1960; 04-09-09 at 06:47 PM.
Reason: clarification - why the heck am I still writin on this?
Bike ≠ Car ≠ Ped.
Next time, copy such a post into a text editor program just in case.
Originally Posted by duffer1960
RicFang, I think you're not finding an answer because you're not using the correct terminology. "Weight" doesn't apply to what you're looking for.
And then here:
If the word "pounds" is used at all, it'll be in "foot-pounds", which is a measurement of Torque.
Big explanation of torque here:
Let's see if I can condense all of this...
Gears and chains are a fancy method of transmitting leverage through a system. Did you ever do an experiment in school with levers and fulcrums? Remember setting the fulcrum point near the object you're trying to lift, which means that your end of the lever is longer. Your end moves farther than the other end, but you can easily lift a larger weight. Move the fulcrum closer to your end, and although you have to push harder, the other end will move farther.
On the bike, changing to a bigger gear in back is like moving the lever's fulcrum closer to the weight you're trying to lift.
Make sense? At least a little bit?
If you searched and read for 2 weeks, you need a new search methodology. You would have searched just this site, you would have found what you need.
For basics, see http://www.panix.com/~jbarrm/cycal/cycal.30f.html
For advanced, see http://www.analyticcycling.com/
Thank you for your precious time for the info!
no problem, the panic site is the most fun as you can do what if's scenarios