the difference between 20 mph compared to 25 mph
 

I know someone knows this. Doesn't it require twice the power output to get to 25 MPH compared to 20 MPH. I know it is 25% faster but it takes more than 25% more. I can't remember where I read this.

BTW I tried searching and couldn't find what I was looking for here.

All you need is a bigger hill.

And use a power calculator to find the answer as it pertains to you. There are more variables than just speed. Coefficient of friction, grade, things like that. Speed obviously is the most important, since it's squared in the calculations.

25^2 to 20^2 = 56% more required to overcome just the increase in the speed, not including the other considerations.

according to this:
http://www.americanroadcycling.org/T...alculator.aspx

at 175 lbs, 176 watts will get you 20mph, and you need 275 to get to 25mph.

If it were pure wind resistance I think you are close, but at low speeds there is more than wind resistance involved.

http://www.analyticcycling.com/ForcesPower_Page.html

There are lots of variables to play with here, but I tweaked a few and had 20mph=309 watts, 25mph=445 watts, so ~50% more.

edit: no power meter, but that seems high. rffff's numbers seem like they would be more realistic. Either way, the % change is similar.

Speed is a linear funtion.
Power required is an exponential function.

-Z

kinetic energy = 1/2 mv^2. Ratio = 25^2/20^2 = 1.5625 more energy required to sustain speed.

And in terms of difference in perceived effort:
20: this burns a little
25: mommy come save me.

My experience (flat terrain, calm to no wind), to maintain speed:

20mph~200W+/-
25mph~300W+/-
30mph~400W+/-

Your results may vary.

Not quite. That's the difference in energy, or the amount of work required to go from 20mph to 25mph. This is different from the amount of power required to sustain that speed, which is what the rider needs to input in to the system to overcome losses like drag and friction.

want some more math? force of wind resistance = .5(mass density of air)(surface area)(drag coefficient)(velocity)^2. at 25 miles per hour you need to resist (25^2/20^2) times as much force than 20 mph

about 100 watts is right.

I'll let the math majors handle the math.

I started road biking 6 months ago. I started out at 210 pounds and rode around 14-16mph. It's been six months now and I'm 187 pounds and I just completed a three hour ride yesterday at 20.1 mph. I was PUSHING it. It was all flat, along the sea and there was very little wind for the entire ride.

I had been around 17-18mph for awhile and then in the past month or two it jumped to 19-20mph for short rides. Averaging 20mph solo for a three hour ride is an accomplishment - for me.

25mph? Are you kidding? I won't see that for years. To me the power and cardio to get there is astronomical. I can push 25mph for extremely short time periods on flats but can't imagine doing an entire ride like that.

So mabe it's 56% more via math, but in the real world, well, it's impossible! (for me)

The power to overcome aerodynamic drag is proportional to V^3. Drag (force) is proportional to V^2.

The first two are OK but the last number tells me either your power meter is out of calibration or you don't spend a lot of time at 30MPH :)

56% may or may not be right, but know that 56% is a HUGE amount of effort. Remember this is a sport where 1% is a big advantage. 56% may as well be the other side of the moon.

This man got it right. The reason for the additional factor is that the higher speed means higher force (v * v) in less time (1/t where t = k/v).
P = f * d / t = k1* v * v / (k2 / v) = k3 * v * v * v.

How well the f = k * v * v as the sole drag component assumption holds is subject to debate.

It's about double.

Power is roughly proportional to velocity cubed:

P2/P1 = (V2^3)/(V1^3)

= (25^3)/(20^3) = 1.95 ~ 2

(P1, V1 are the power and velocity respectively at 20mph. P2, V2 are the power and velocity at 25mph)

Agreed. The first two sound right but the third is low. I'll have to check PerfPro but I'd guess it's well over 500 maybe 600.

NO. Please stop perpetuating this. It's wrong.

Just because an equation has an exponent in it doesn't mean that it's an exponential function.

Power is a cubic function of speed, meaning P=a*v^3 + b*v^2 + c*v + d. Finding a, b, c, and d is sometimes difficult, but at high speeds, the v^3 term is much more important than the other terms, so they often get ignored, and rightly so.
Moral: going twice as fast requires 8 times as much power.

Force is a quadratic (square) function of speed, so F = a*v^2 plus the lower order terms.
Moral: Going twice as fast requires 4 times as much force to the pedals, but of course you have to turn those pedals twice as fast, hence the 8x from the power equation.

An exponential function is one where the variable in question IS the exponent. Completely unrelated to cycling, but bank interest, bacteria growth, and radioactive decay are all exponential functions with the form Y=a^x where (a) is a constant and (x) is a variable.

brianappleby: It's obvious it's not the brightest of math stars who are looking at this thread. You are, of course, absolutely spot on in the explanation. I'm just worried the effort is wasted.

Suffice it to say:
1) Power is a function of velocity cubed (not squared, not exponential)
2) to go from 20mph to 25mph, you require about double the amount of power.

I wonder what would happen if you threw a copy of Newton's Principia into the monkey house at the zoo.

no need to waste a good book: we just saw what would happen in the exchanges above

actually, i take that back, if you have large enough a population of monkeys and give them enough time, some of them might just be able to come up with the laws of Newtonian physics. Here, Newtonian physics just get misinterpreted.

Basically, this.

Force is the time derivative of momentum and momentum is mass times veloxxzsz aodhh *%%& dxxz To be or not to be that is the quesndsdlllddn vcv [EndOfTransmission]

So close yet so far away. :cry:

Really?