|suprising formula for power on a bike||Dog|
Jan 2, 2002 11:57 AM
|Got this from an email list; does it seem right?
There is a very quick sum which will give you a surprisingly accurate estimate of power when climbing a hill.
Simply multiply your speed in miles/hour by the gradient and your weight in lb ( weight of you + bike).
If you multiply the answer by two you get your power in watts!
In other words: Power in watts = 2 x S x G x W
where S= speed in mph
G = Gradient (ie if its 1 in 10, multiply by 1/10 )
W = your weight in pounds.
So if you are climbing at 6mph, up a 1 in 10 hill ( 10% ), and weigh 200lb ( you + bike ) you get:-
2 x 6 x 0.1 x 200 = 240 Watts.
If you have studied physics you may not believe me - the units are all mixed up! It is correct within 1% though ....
Here is the derivation.
The physics for this are fairly straight forwards. When climbing at slow speeds (ie less than 10mph), wind resistance may be ignored. If the hill is quite steep ( around 10%, ie 1 in 10 ) rolling resistance is not too important either.
Please excuse my old fashioned units. They are English ones as I am English!
If you want to impress your friends, use the Horse power version, or if you are scientific by inclination, use Watts!
Power = Distance x Force per unit time.
Power = Foot Pounds per second.
OR = Pounds x Feet per second.
Now the force you need to apply depends on Weight ( You + Bike ) and gradient.
If for example the hill is 1 in 10, or 10%, you would need a tenth of the force you would need if you were going vertically.
Me plus bike weigh about 200lb (no I'm not too proud of that!)
Suppose I'm climbing a 1 in 10 hill at 6mph, which is 8.8 feet per second ( tough going if its a long one! )
Then power = Pounds x feet per second x gradient
power = 200 x 8.8 x 0.1 = 176 foot pounds per second
Who wants to use those units??? How about using Horse power!!
One Horse Power = 550 ft lb per second,
So my power in climbing a 1 in 10 hill is 176/550 HP = 0.32 HP
So I'm climbing the hill at a third of a horse power! Sounds impressive ....
If you prefer watts,
One Horse power = 746 Watts.
0.32 x 746 = 238 watts .
And that is not so very far from the original estimate! It works because if you multiply the conversion factors for feet/ second to mph by the other conversion factors you get about 2 :-
88/60 x 1/550 x 746 = 1.989. Strange but true!
You may not want to bother to work all that out as you go along ( though it helps to pass the time! )
So a "difficulty factor" is obtained by just multiplying the speed buy the gradient:-
10mph x 1/10 = 1 is the standard -
same power as 5mph up a 1/5 and so on.
It only works well for 10mph or less, as at high speeds you will be using a lot of energy overcoming wind resistance. For example, at 19mph wind resistence accounts for 150 watts. Wind resistence rises by a huge power law at higher speeds still.
So a difficulty factor of 1 might be the best you are able to achieve for a short period, and tackling hills at a factor of 0.5 might be a sustainable level for several hours, for example. Its a relative measure for one individual as weight is not included. I've found it very useful when tackling a 1/6 : it stops me trying to go too fast!
Incidentally a "difficulty factor" of 1 is about half a horse power!
Jan 2, 2002 12:18 PM
|I have a favorite climb. Here are the details:
Based on data from analyticcycing.com I estimated my wattage to be about 285. With this formula it is 227:
2 x 11.7 x .0625 x 155 = 226.6875
I believe that this formula looks attractive, however it will be increasingly innacurate as S (speed) increases because it can't account for wind resistance. I would suspect that this formula is actually accurate at 8mph or less, not 10mph as you stated. Either way, it is a fairly easy calculation when you don't have analyticcycing.com handy.
Personally I can't wait to get my Polar power kit. That will make this stuff much easier. ;)
By the way, what email list did you get it from? I have been contributing and lurking on email@example.com for about a month. It is a very serious minded list with some great insights on training for sustained power.
Jan 2, 2002 12:24 PM
|Yes, this is something you might actually be able to do in your head, assuming you're not climbing too hard
|Speed factor||Kerry Irons|
Jan 2, 2002 5:19 PM
|The reason this works is because you are simply calculating the effort to lift a weight. At low speeds, nearly all of your effort goes into overcoming gravity, so you can ignore aero drag and friction. As the gradient gets smaller, this becomes less and less accurate. A 10% grade is pretty steep, so nearly all of your effort is fighting gravity.|| |