Is the formula or conversion for how much a person’s weight effects his/her work and power output on the bike? For example, Rider X weighed 170 lbs. and dropped 15 pounds. How much more work and power would he put out? Obviously there are other factors, but I was wondering if there was anything standardized out there.
Is the formula or conversion for how much a person’s weight effects his/her work and power output on the bike? For example, Rider X weighed 170 lbs. and dropped 15 pounds. How much more work and power would he put out? Obviously there are other factors, but I was wondering if there was anything standardized out there.
On the flat, weight only plays a role in affecting rolling resistance. The faster you are going the less important it becomes as a variable. In most instances it is a tiny contribution to the whole. go to analyticcycling.com and put in osme numbers and you will see it is seconds over an IM distance. 15 lbs might save you a minute.
Hell Frank, I should have thought to drop you a line. But I’m looking for a formula or conversion though. I’m making up a lab for my science class. It’s also a way I can buy more shit and write it off my taxes. Including PCs. Any educators discounts?
I’m confused. Why do you think that dropping weight will increase power output and work done?
I believe analyticcycling has the formulas they use available to all.
Sorry, the only educator discount is for those educators wlling to study them.
I don’t know that for sure and if I’m wrong someone will correct me.
Remember the basic formulae.
F=ma
W=md
P=Fv
Per the above, to achieve the same acceleration, less force (and power) is required as mass decreases, and less work will be done. on flatland, power is a result of drag forces (aero and frictional) and velocity. Mass is only important as it pertains to the change ion those forces, which will also decrease (generally speaking) with a decrease in mass. The magnitude of reduction is also quite small.
I realize all of that. But the basic formula for power is P = W/t.
P=W/t is another way of saying P=Fv, isn’t it. mass is still in the numerator, not the denominator, so the relationship between mass and power / work doesn’t change.
I realize all of that. But the basic formula for power is P = W/t.
then what’s your question?
I did the calculations way back when, and boiled it down to this rule of thumb: For any serious climb, 8 pounds cost you one minute for every 20 minutes of climbing, given equal power output. That’s why TDF cyclists are so obsessive about their weight.
I think what you’re trying to say is how much more work would the heavier rider have to produce to keep up with the lighter rider, or how much faster would the lighter rider be without the weight. It sounds like you’re saying the rider gains power by losing weight, which are two completely different aspects.
"On the flat, weight only plays a role in affecting rolling resistance. "
Unless you assume that the more massive body is probably larger as well, which means more aerodynamic resistance. I believe this is a bigger affect than most people give credit for. It takes a lot more power to push the shoulders, legs, head, chest, and arms of a lean 200lb guy through the air vs some scrawney Ivan Basso type.
"On the flat, weight only plays a role in affecting rolling resistance. "
Unless you assume that the more massive body is probably larger as well, which means more aerodynamic resistance. I believe this is a bigger affect than most people give credit for. It takes a lot more power to push the shoulders, legs, head, chest, and arms of a lean 200lb guy through the air vs some scrawney Ivan Basso type.
I would agree but this is not strictly a weight effect. Actually, there is another element that would slow you down overall despite the same average power output and it is ignored by all the calculators. It is the energy required to constantly speed up the bike after the wind slows it down 180 times a minute or so from the irregular application of pedal force we all have. The more irregular the application of pedal force (more of a “masher” or a lower the spinscan number) the bigger this effect will be.
Frank, you never cease to amaze me. Just when I think the Repubs are the Masters of Spin, I come onto ST and see you hard at work, and realize they are rank amateurs, utter posers, when compared to you.
<< there is another element that would slow you down overall despite the same average power output and it is ignored by all the calculators. It is the energy required to constantly speed up the bike after the wind slows it down 180 times a minute or so from the irregular application of pedal force we all have. The more irregular the application of pedal force (more of a “masher” or a lower the spinscan number) the bigger this effect will be. >>
There’s probably a reason all the calculators ignore this mystery “element”, perhaps several:
-
it doesn’t exist
-
if it does, it’s probably so small as to beyond measurement (ie: smaller than the margin of error in the calculations)
Hey, Frank - do you happen to know of a product I could buy (for about $800 or so) that would help remedy this non-existant problem by any chance??? 
I would agree but this is not strictly a weight effect. Actually, there is another element that would slow you down overall despite the same average power output and it is ignored by all the calculators. It is the energy required to constantly speed up the bike after the wind slows it down 180 times a minute or so from the irregular application of pedal force we all have. The more irregular the application of pedal force (more of a “masher” or a lower the spinscan number) the bigger this effect will be.
Could you calculate this for me? Assume 40k distance, CdA = 0.25. Crr = .005. 90kg bike+rider. Assume spinscan number of 70. How much slower would that guy be than one who pedals at spinscan = 80?
Let’s make it easier. Assume zero power is applied through 1/6 of each pedal circle. At 90 rpms, that is .11 seconds of no power, I think (you should check). In that amount of time, how much does a bike going 9.0 m/s slow down?
You’ll need to first calculate the total inertia of bike, rider and wheels.
Since the effect you describe is so very important, please tell me how big it is. I need an exact figure because I’m trying to decide whether to buy thinner bar tape to make up the difference. Or, I might trim my fingernails to remove some weight. That ought to even things out.
It does exist and it is not trivial. I suspect it has just never been considered by those who do cycling calculations. If it has, I would like someone to direct me to a calculation that shows it has been considered and is trivial.
Check out the “for Paulo” thread and this link: http://sportsci.org/2006/ssrowing.pdf
Check out the slides on page 40-41. The math is identical except we are dealing with air resistance, not water resistance.
When rowing power is applied only when the oar is in the water and during that time the boat is accelerating. The rest of the time the boat is “coasting” and decelerating. The average power would be about the average power while the oar is in the water divided by about 2 (depends on the relative lengths of the two cycles but it is about 2).
From the slide on page 41:
How much speed could be gained by
reducing velocity fluctuations by 50%?reducing velocity fluctuations by 50%?
• Estimated ~5% efficiency loss due to velocity
fluctuations (see Sanderson and Martindale
(1986) and Kleshnev(2002)
• Reducing this loss by 50%would result in
a gain in boat velocity of ~ 1%or ~4
seconds in a 7 minute race.
• Sliding rigger effect probably bigger!
due to decreased energy cost of rowing and
increased stability (an additional 1%+ ?)
Go to this page at analyticcyling.com:http://www.analyticcycling.com/PedalForcesAtPedal_Page.html
The only forces that actually drive the bicycle are the green forces. And the average force would the the average of all the forces. When we combine both legs together we will see that about half the time the power is way above the average and half the time it is way below (although it is never zero as in the rowing model). The math is the same. for the above reasons, I would expect the severe masher is seeing somewhat less than 5% loss calculated for the rower, say 4%. So, let’s say 3 seconds in a 7 minute race can be gained. If we extrapolate that to a 300 minute race (5 hour bike split) that would be 128 seconds (over 2 minutes I believe that is) saved. Not large, but not trivial. And, if one can do it it is “free speed”.
If my math is incorrect I am sure you will show us all the steps involved in making this calculation and correct me (and those poor misinformed rowing scientists). I have actually done this calculation, by the way although it was a long time ago, and the above simple analogy gives the same ball park answer I calculated. If I am wrong just don’t tell me I am an idiot. Show me where I went wrong.
And, if you are looking for a product that might be able to help you smooth out your pedal stroke (although then it won’t quite be “free speed”) I believe I do know of one. 
Cheers.
I have done this calculation, it is really not very hard using a spread sheet and you can change all the variables if you want, cadence, mass, aerodynamics, etc. I am not going to do it again for the purposes of this thread. It is easy enough if you understand the math involved. Do it yourself.
That being said, as I remember correctly. The main issue here is not the mass of the rider or the cadence or anything else but the variation in the force around the circle. Actually, a heavier bike rider combination is better for this because the speed variations will be smaller with more mass but it is a tiny issue because the rider is huge compared to the bike.
The problem is that the wind resistance varies with the square of the velocity. Any variation in speed must mean that the average wind resistance is greater than if the force application were constant and the wind resistance were constant. This slows the rider down, it has to. The more the variation, the more it slows you down. The only question is the magnitude. Probably a little less if you have good aerodynamics than if you have bad. And a little less if you are heavier rather than lighter. But, it will slow you down, none-the-less. Probably, as I calculated in another answer, about 4 minutes for an IM bike split, but since you can’t make this perfect one can probably only gain 2 minutes or so improving this aspect of their game.
I might add since this since I already referenced this page:
http://www.analyticcycling.com/PedalForcesAtPedal_Page.htm
If you don’t believe that it is possible to achieve huge power increases by changing your pedal stroke look at all the red arrows. Those are forces on the pedals that do you absolutely no good. It is energy expended that is completely wasted. If one really wants to get faster one needs to learn how minimize the red forces and maximize the green. This requires absolutely no more energy from the rider, just a different coordination to direct the forces differently.
In my opinion, this is where we will find the major contribution of the PC effect, how the large increases in power seen by new users are mostly accounted for once these studies are done.
If you don’t think such changes can be made or that making such changes would not result in large power increases I would love to hear your analysis as to why.
Cheers
It does exist and it is not trivial. I suspect it has just never been considered by those who do cycling calculations. If it has, I would like someone to direct me to a calculation that shows it has been considered and is trivial.
.
This is another example where you have made a claim, refuse to provide evidence to support your claim, and insist that everyone else prove you wrong.
Come on! Show me the numbers!!!