AdelaaR

KATCHING!
This thread just got more interesting

achoo

Bingo.

On level ground, even though the force of gravity is the same there's no change in vertical distance so there's no work done in the same (or opposite) direction of the gravitational force. (There's still going to be work done to overcome drag and mechanical losses.)

If you're gaining altitude, though, you need to do work against the force of gravity to gain that altitude, and that takes energy and thus power.

If you're descending, the force of gravity, for lack of a better word, does work on you and you go faster. Or heat up your brakes.

BoSoxYacht

I made the comment to a Mod, and he thought it was funny too. I guess another Mod thought otherwise.

achoo

In general yes. But given the conditions set forth, no.

The general unbounded case of any wind, any speed, any cadence is somewhat difficult to solve for....

AdelaaR

I believe we have overlooked something:
If you apply more torque to the pedals, you will squish the front tire more resulting in more loss in hysteresis energy.
Or shall we assume, for simplicity's sake, that the rider is pedaling in perfect circles?

achoo

There are other kinds?

AdelaaR

Most definitely.

MikeyBoyAz

In medical terms, this thread has started CTD = [circling the drain]. Sad.

sreten

Hi,

It is always true. 28/18 times harder on the pedals to get the 28/18 extra power at the same
cadence. Your points have nothing to do with the veracity of the statement your commenting on.

rgds, sreten.

28 mph no wind vs 18 mph into 10mph, both on the flat,
drag = 28mph, ignore everything other than drag.

mprelaw

Was that before you needed at least a bachelor's degree in physics to read the threads?

Dudelsack 

My wife had a hysteresis about a year ago.

rpenmanparker 

sreten you are off base. You can set up a situation with same cadence, same force, different gearing and different bike velocity that gives different power due to the velocity difference. That is what the 18/10 and 28/0 situation is. Not necessary to get more force at the pedals to get the higher velocity in the absence of the headwind. Therefore not necessary to have more force at the pedals to get the higher power generation. I have been paying attention to this ridiculous thread I inadvertently started. Have you?

sreten

Hi,

It is you lazily not paying attention to the original post that
I commented on the reply to. I comprehensively understand
the physics and don't need telling such waffle as the above,
which reading it through is nonsense, if intended to follow.

rgds, sreten.

BoSoxYacht

BTW, you caused me to spit out beer via the nasal passages.

rdtompki

Perfectly said, but let's not forget you can have some fun on the way down.

rpenmanparker 

Ignoring minor influences and concentrating only on wind it takes the same force to ride 18 mph into a 10 mph headwind as to ride 28 mph with no headwind. But to keep the same cadence at a slower speed you have to gear down. So as the guy who originally brought this up asked, you have same force, same cadence, different gears and also different power because same force but different velocity. He asked would you notice the differences. The answer is only the speed of scenery passing you by and sensation of speed. What are you arguing about?

sreten

Hi, I'm not. I told him he was very wrong. He now agrees. rgds, sreten.

The pedalling force is half and of course you would notice that.

rangerdavid

this thread has true 41 potential

Brian Ratliff

Eh? Are you forgetting that the drivetrain is a torque converter? You have same force at the wheel, same cadence, different gears. Because you have different gears, to have the same force at the wheels, you necessarily have proportionately different force at the pedals. You know you are going 18mph into a 10mph headwind rather than 28mph into still air because you are pushing the pedals less hard.

FWIW, the speed you'd have to be going into a 10mph headwind to equal the power needed to roll 28mph in still air is ~23.5mph. Bonus points for anyone who manages to verify my calculation. Hint: it involves solving a quadratic equation...

(Double bonus points if you can remember the quadratic equation without looking it up. I didn't get double bonus points... )

EDIT: okay, I should say my calculation only works if you assume drag force is proportional to velocity. It's not (force is proportional to the square), but at bicycling speeds, the error shouldn't be too bad.

EDIT2: if you believe the online cubic equation solver, the real answer is 21.8mph. About 10% error using a linear assumption rather than the quadratic for the force/velocity relationship.

njkayaker

There's a limit to that. And you are a bit late.

njkayaker

No, still wrong. And your ad hominems don't change that.

You are still moving when track standing and so doing work.

Brian Ratliff

I think we are past this now. I mean, you specifically quoted an authority you apparently trust which states my point precisely. Brandt is an abrasive fellow, but he is generally a good engineer. I thoroughly enjoyed his book on bicycle wheels.

rpenmanparker 

Of course you are right. I keep only seeing half of the story. So let me try to extrapolate the learning. I won't propose this as being right, but rather ask if it is to see if I understand now: At the same cadence but different gearing the different powers generated by the identical velocity of motion of the rider's legs X the different forces at the pedals is the same as we you would calculate by multiplying an identical force on the rear wheel X the two different bike velocities. When the cadence is the same, it is the force that varies at the pedals and the leg velocity that is the same, but for the bike it is the force that is the same and the velocity that is different.

Thanks for your patient explanations.

rpenmanparker 

Damn, of course, as always you are right. Even though I know that in advance, something about the way I read your explanations just doesn't allow me to understand what you are saying. My apologies. Thanks for staying civil despite my foolishness.

Brian Ratliff

^^^
yes, in the special case we are discussing here (28mph in still air vs. 18mph in a 10mph headwind). The force at the wheel is constant. If the drivetrain is adjusted so the cadence is identical, then the speed of the wheel differs in direct proportion to the difference in pedal force. The ratio between the front chainring and the rear cog describes this relationship exactly.