Naked Science Forum
Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: Cheese2001 on 12/10/2012 23:57:36
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OK, science folks...
I enjoy riding my bicycle with an infant seat for my son on the back. I weigh about 200 pounds, my son is about 30 pounds, and the bike with seat is around 40 pounds. There are several hills in the area around my home, and while climbing one, a thought struck me. My body connects to the bike at about four places, one foot on each pedal and one hand on each side of the handle-bar. I was in a very high gear up a steep hill, but my legs were still tiring. What I'd like to figure out how to calculate is how much force I was demanding from my thighs with each pedal? It much be less than my body weight, as my downgoing foot stayed attached to the pedal, weight was still transferred from my hands to the top of the handlebars. I have regular pedals, not click-in shoes, so the up-going foot is as light as possible. Any ideas how to work the kinematics to figure it?
If I maintain a constant speed going uphill, then the force out of my thighs should be the same as the portion of the weight vector parallel to the slope of the hill plus friction losses?
Thanks for any ideas!
-Cheese
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You have to measure the mechanical advantage of the whole system (http://jakirkpatrick.wordpress.com/2012/02/12/which-way-round-do-the-pedals-go/). That's the ratio of the crank length (http://www.nettally.com/palmk/crankset.html) divided by the outside diameter of the rear tire times the gear ratio. That will give you the ratio of torque on the crank to torque on the rear wheel. Tire sizes are the approximate outside diameter, typically 26" for adults.
The force on the pedal varies with the angle of the crank. When the force is vertical and the crank is horizontal, you get the maximum torque from a constant force. The torque is then equal to the force times the crank length. Longer legs do better with longer cranks. A longer crank requires less force on the pedals. If the crank is too long, your bottom pedal may drag on the ground when you lean in a fast turn. (If you have elliptical sprockets (https://encrypted-tbn1.gstatic.com/images?q=tbn:ANd9GcRB8oxV17y037vw4E5ti3rZJ7r_6RaW0X2Ee7_hcJlHS14JV6I4), they reduce the importance of the crank angle. They increase the mechanical advantage when the crank is close to vertical.)
You use a low gear uphill, and a high gear going downhill. If you have a gear box, count tire revolutions while turning the crank thru one revolution. To calculate the gear ratio of a derailleur system, count the teeth on the front and rear sprockets, F and R. The ratio is F/R. Your lowest gear uses the smallest front sprocket and largest rear sprocket. In this image, I count 40 teeth on the front sprocket and 28 on the rear sprocket. So the ratio is 40/28.
(https://www.thenakedscientists.com/forum/proxy.php?request=http%3A%2F%2Fwww.globalspec.com%2FImageRepository%2FLearnMore%2F20123%2F748px-Derailleur_Bicycle_Drivetrain.svg9d0b480ee02b4859baaf2613c25d9a00.png&hash=71ce8e467816c014436b350fc14200a7)
With 270 lb total on a 10% grade, you have 27 lb of tangent force on the tire. With a 26" tire, that's about 702 inch-pound of torque. With a gear ratio of 4:3, you need 702 * 4/3 = 936 inch-pound of torque on the crank. With a 6" crank, that's 156 lb average force. You probably would not be able to get up that hill because your 200 lb weight can only be applied perpendicular to the crank shaft for a small part of the time. You would do much better with a 1:1 gear ratio and a 7" crank. An elliptical chain ring might also help on those steep hills.
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Phract - nice in depth answer, but I am not sure about your figures at the end; I can go up a (short) 10pct hill in a high gear (much larger than 4:3) without getting off the saddle. I find it hard to believe that I am using a force of those magnitudes.
nb Most adult road bike tyres are about 26-28inches diameter - you seem to be using it as radius