Rider weight, power, hill, speed

Peddle Up!

I suppose I ought to able to dredge up my school physics and work this out for myself, but I'm feeling particularly idle tonight so I'll ask to see if anyone else has done it.

Is there a formula that relates the above so that the speed a rider of given weight, capable of outputting a specific amount of power, on a given upwards incline can be calculated, albeit approximately?

What I'm interested in is the effect a bike plus rider has on climbing in a measured rather than anecdotal way.

I appreciate that there are other important variables and caveats but I hope you can see what I'm getting at. If not, I'll just go back to sleep.

nickel

Could try this website: http://bikecalculator.com/wattsMetric.html

desweller

For a given weight and climb, a rider with twice the power of another rider will complete the ascent twice as quickly, ignoring all other factors. For a power and ascent, a rider + bike of half the weight of another rider + bike will complete the ascent in half the time.

Potential energy = altitude x mass.

'Power' however, is one of those terms that covers a broad topic. If one rider produces peak power of 300W at 90rpm, but experiences a 200W dropoff 10rpm either side of his/her peak power point, they'll struggle compared to a 250W rider who can remain within 50W of peak power 10rpm either side of their peak power point.

Note that my figures are completely hypothetical, I don't have access to power data but I do work with engines and the methods for quantifying performance are similar!

siamon

There are some equations on this thread that might give you an idea how complicated it gets:-

viewtopic.php?f=40011&t=12833196

citrus_

It can be simplified to be quite easy actually, for all variables constant:

Power = Velocity x Force

According to Newton, "Every action has an equal and opposite reaction," and so the force is equal to the gravitational force that you are overcoming, equal to m x g x sin(i)

So:

Power = Velocity x Mass x g x sin(i)

Where velocity is your speed in metres per second, mass is the weight in kilograms of rider plus bike, g is about 9.8 for riding on the earth, and i is the incline in degrees. Ie for a 1 in 4 hill, i would be 22.5. This outputs the power in watts.

Obviously this is a crude way of doing it as it ignores other forces, the most important of which being friction, which makes a massive difference

Pross

From a slightly different angle I was thinking the other day that a neat little app would be something where you could input the vital statistics of you and your bike and then see what change in power would result from swapping components (or losing body weight) at given speeds and gradients (with all other things being equal). It would help answer all the questions you get about the best upgrades be a nice project for an engineering undergraduate.

citrus_

Pross, you could kind of do that from my formula, if we rearrange we can see that velocity is proportional to 1/mass, which basically means that if you reduce mass by about 1%, you can increase your velocity by about 1%

bompington

Easy enough to calculate that, say, spending £200 more on a carbon stem to reduce weight by 10g will therefore increase my speed by about 0.001%, or about 0.006 seconds over a 10 minute climb.
On the other hand, if I cut out the biscuits and got my weight down to 12 stone the time saving would be more like a minute and a half :oops:

desweller

I have often wondered how well different riders cope with the deceleration between strokes. I guess this is where the whole 'pedalling in circles' mantra comes in.

sub55

citrus wrote:

Pross, you could kind of do that from my formula, if we rearrange we can see that velocity is proportional to 1/mass, which basically means that if you reduce mass by about 1%, you can increase your velocity by about 1%

don`t know about that :!:
if you reduce mass by 1% your velocity would increase but so would the drag. (the power required is equal to the cube of the speed) so velocity wont increase by 1%.

MartinB2444

"Power = Velocity x Force"

Does this still apply when going up hill? When stationary, assuming stabilisers in place :shock: , you are having to apply force to avoid rolling down hill, that must require a certain amount of power. So doesn't gravity have to be factored in.

desweller

MartinB2444 wrote:

"Power = Velocity x Force"

Does this still apply when going up hill? When stationary, assuming stabilisers in place :shock: , you are having to apply force to avoid rolling down hill, that must require a certain amount of power. So doesn't gravity have to be factored in.

It gets complicated when you're dealing with muscular systems. Viewed externally, no power is required to maintain a state of equilibrium (think handbrake rather than hill start).

MartinB2444

DesWeller wrote:

MartinB2444 wrote:

"Power = Velocity x Force"

Does this still apply when going up hill? When stationary, assuming stabilisers in place :shock: , you are having to apply force to avoid rolling down hill, that must require a certain amount of power. So doesn't gravity have to be factored in.

It gets complicated when you're dealing with muscular systems. Viewed externally, no power is required to maintain a state of equilibrium (think handbrake rather than hill start).

Ok, but the second you start to roll, however slowly there is a certain amount of power that has to be used to overcome gravity. In the case of a Saturn V rocket I guess it was quite a lot of power needed just to make it "weightless".

citrus_

sub55 wrote:

if you reduce mass by 1% your velocity would increase but so would the drag.

Going uphill, drag is going to be pretty trivial, its the friction you've got to worry about, and I did put a disclaimer about that in my original post!

citrus_

@MartinB,

Yes the formula is all correct for going up a hill. The force part is the bit that takes into account overcoming gravity. The force parts that I have ignored however are those overcoming drag and air resistance, which may well be very significant.