timmyquest

What?! What in the world are you talking about?

timmyquest

It's happening...don't say i didn't warn you.

bdcheung

He's being pedantic and splitting hairs. Ignore the "BUT IF YOU REMOVE THE CRANKS YOU AINT GOIN NOWHERE" posts.

timmyquest

I can't. Removing the cranks was my idea because their counter to my argument about mass was "Well what if you cut off your foot, if all mass is the same. No foot is less mass but i bet you go slower". So i ask if lighter is better, lets remove a crank. I was half joking...but not really. I'm not kidding, they really seem to believe this crap.

Phantoj

times a million... a = 0 for constant velocity by definition. a = dv/dt, as timmyquest posted previously.

If you're traveling at a constant velocity, on a flat road, you're pedaling!

bdcheung

I'm making this nice and big and red so nobody will miss it:

There are always forces acting on the rider+bicycle system. Even at a constant velocity on a flat road, you are being decelerated - that is, accelerated in the negative direction - by air and rolling resistance. Which is why, if you stop pedaling, you coast to a stop (if you go from a speed of X to a speed of zero, then your speed is decreasing over time, which means you are accelerating in a negative direction).

Which means, in the F=ma equation, even on a flat road (unless you're riding in a vacuum with frictionless tires on a frictionless road, which is to say the least impossible), acceleration is NEVER zero.

KyleOndy

^ For the Win.

bdcheung

Read what I just wrote in big red letters.

timmyquest

Wow...

K&K_Dad

this is all getting to complicated.. I'm gonna start studying string theory..

timmyquest

Why don't you just say "Hey, i think i was kind of wrong. I think i learned something." It wont kill you, and i'm still willing to do the same.

calRider

Actually, you're wrong here. Once all of the forces are in balance (i.e. constant velocity), there is no acceleration.

I will agree that you're never in a state of constant velocity in a bicycle, simply because no one can generate perfectly constant power.

bdcheung

Stop overthinking it.

Yes/No question: if you're on a perfectly flat road, going 20mph, and you stop pedaling without changing anything else about your body position or bike, will you eventually stop?

timmyquest

There is NEVER A TIME in the real world that all forces are in balance and a=0. It is impossible. Just because your speedometer says 17mph for 2 seconds does mean you weren't being accelerated.

Phantoj

if v is constant with respect to time, what is dv/dt?

bdcheung

100% correct.

But even at constant velocity there are forces acting to decelerate you and you are counteracting those forces by acceleration via pedaling.

Phantoj

If you stop pedaling, the vector sum of the forces is no longer zero.

bdcheung

in theory? zero.

in reality? DNE since v is never constant as was pointed out previously.

timmyquest

It never was zero in the first place.

bdcheung

i think we've got five people arguing two different ideas:

1) At a true "constant velocity" a = 0. This is true. However, v is never constant in real-world applications.
2) Since v is never constant, you're always under some sort of acceleration.

timmyquest

And therefore effected by M...

Even when dealing with the aerodynamics problem because p(momentum)=VM

bdcheung

only if a isn't perpendicular to the direction of v

(i.e. only if you're going up, or down, a hill. and even then M is overshadowed by many of the variables affecting a, since a is a composite of not only gravity but also air and rolling resistance)

calRider

if you are moving at a constant velocity, then dv/dt = 0...wind resistance, friction...none of that stuff matters.

Phantoj

OK, in the "real", rather than theoretical, case, you can say that each press on the pedal causes an acceleration, and each dead spot causes a deceleration, but then you end up figuring that heavier bikes are faster because aero drag is nonlinear, so a more constant speed is more efficient.

dunningrb

No, it's correct. Constant velocity entails zero acceleration. And you won't coast to a stop if you stop pedaling going downhill.

An F = ma analysis is much more complicated than one based on power and energy, which is why we normally avoid Newton's second law when talking about bicycles. But if you want to go in that direction, you need the external forces acting on the bicycle:

(1) The contact force at the front wheel and road, pointing backward.
(2) The contact force at the rear wheel and road, pointing forward if you're pedaling and backward otherwise.
(3) Air resistance, the direction of which depends on your velocity relative to the air (in a simple model).
(4) Rolling resistance, the most difficult of these forces to model. In a simple model, the magnitude depends on total weight, the grade of the road, the type of tires, and the quality of the road surface. Points backward. There's a large margin of error in this term.
(5) Gravity. Points down.

To get the acceleration out of this model, we need to know such things as the total mass of the bike + rider and the aerodynamic profile of the bike + rider. If we want to be really detailed, we need to know how the frame is flexing when you pedal. (I don't think anyone really models this.)

In principle, we don't need any of the "internal" forces, such as the force applied by your foot to the pedal, because all of these cancel in pairs due to Newton's third law. In practice, we probably need to measure or estimate these in order to work out the contact forces above, (1) and (2). This will lead to estimates or measurements of various moments of inertia for the rotating parts, and the torques applied to those parts.