# Bicycle momentum

1. Nov 4, 2008

### lmace

hello to everyone this is my first post on this website but have used it before and found existing posts and answers very helpfull. i am a engineer at bourneouth university studying BSc Computer aided product design, i am currently working on research into a bicycle braking system to improve the users use of the front brake. Even though the front brake is the most efficant of the brakes users are relutant to use it as they are scared to use it because of the danger of flipping the bicycle.
i would like to be able to work out the mimium speed a user would have to be travelling at to allow the momentum to overcome the reaction force of the users arms and center of mass, to force them over handle bars by angular momentum of the locked wheel.

any help would be very much appriacated.

aplogies for grammer/ spelling i am a designer not english student.

regards luke mace

2. Nov 4, 2008

### uart

Ok the question seems relatively straight forward so I'll answer it first. But after that I'd like to suggest why fear of going "over the handlebars" is not necessarily the main concern with front brakes on a bicycle.

First the question. I'd say all you need to do is to find the change in PE to raise the center of mass of the bike and rider to the "flipping point" and equate that to the initial KE to solve this.

Assuming that the center of mass of the bike and rider is at a distance "a" above the front axle and distance "b" to the rear of the front axle then equation I get is :

$$1/2 m v^2 = mg ( \sqrt{a^2 + b^2} - a )$$

So

$$v_{\max} = \sqrt{2g( \sqrt{a^2 + b^2} - a)}$$

Personally I think that the loss of steering that you get if the front wheel locks up is more of a concern than the flipping issue. As you know the steering on a bicycle is an essential part of how you maintain stability, with the rider continuously making minor corrections to the steering in order to keep his center of mass correctly positioned and therefore the bike upright. I know from painful experience that it only takes a momentary loss of steering (due to front wheel lock) when descending on a loose surface before you can end up in a unrecoverable position and the bike goes down.

Last edited: Nov 4, 2008
3. Nov 4, 2008

### rcgldr

Usually the brakes on a road bike aren't strong enough to lock up the front. For fast braking, I simply raise myself up a bit and shove the bicycle forward so the seat is under my lower abdomen, moving the center of mass back to prevent flipping over. Mountain bikes on the other hand, at least the ones with disc brakes can lock up the front tire.