Recent content by gearhead

yeah, thanks. funny story.

allright, thanks for clearing that up for me, another question: could a similar system work in a hybrid? For example, let's say that you have a generator that produces 5hp at highway speeds to recharge the batteries. 5hp to a 100hp engine is not that much extra power for it to output and...

yes i realize that it's a net loss, but you'd still probably get at least 40 miles or so on it before batteries died, right?

I'm pretty sure someone somewhere has already built one, I've gone over the concepts involved and it does seem both practical and possible. The idea is this: electric-powered car that is recharged by several wind-generators(essentially fans on the roof) that recharge the batteries once the car...

Think about it: if a bicycle is riding towards a car at 30mph, and the car riding at the bike at 40mph, they're coming at each other at 70mph! whereas, if the bike is moving the same direction as the car at 30mph, and the car at 40mph. the car is coming up on the bike now only at 10mph, giving...

They will both accelerate at the same speed, assuming they both have the same coefficient of drag and frontal area. As previously stated: everything falls at the same rate.

The only time during that 4-stroke pedal cycle that 100%(disregarding mech. loss) of the rider's effort(force applied to the pedals) is being directly converted into torque is at 90 degrees and 270 degrees. this is because it is only at these two points that the pedal force is directly tangent...

As far as determing how much torque ends up on the rear wheels, just take that torque developed at the crankset and multiply it by the gear ratio. For example, let's say that you're pedaling up a steep hill out of the saddle: you're exerting your body weight, let's say 600N's on the 175mm...

How much torque you end up getting on the rear wheel is directly dependent on solely the gear ratio and the crank arm length. The smaller the gear ratio(ex: 0.5:1), the less torque you have on the rear wheel. the larger the gear ratio(ex: 2:1), the greater torque you have on the rear wheel...

At 90 and 270 degrees, the force of that the rider is exerting onto the pedals is *tangent* to the crank.

allright, thanks, it makes sense now

wait, but how can the magnitude of torque be the same for both the hub and the wheel? the hub and the outside of the wheel act as a lever system. since force is being exerted on the hub, and the hub is approximately 13" away from the outside of the wheel, then the force on the outside of the...

okay, so then you are suggesting that the friction between the outside of the wheel and the ground is opposing the motion of the wheel? (you are suggesting that there is a static coefficient of friction to be broken?) there is torque on both the outside and the inside hub of the wheel, it's...

allright,just to make sure, so then the torque i calculated in that equation above is the force that is exerted on the OUTSIDE of the wheel right?

allright, so then my question is, at what point on the wheel is this force exerted? Is it exerted on the outside of the wheel where the tire is, or is it exerted on the inside hub of the rim?