jshort2010

There is formula for on wiki article "Bicycle performance" known as

P = g * m * V_g * (K_1 + s) + K_2 * (V_a)^2 * V_g

P - Power in watts
g - gravity
m - mass of the rider plus bike
s - grade of surface
V_g - ground speed m/s
V_a - air speed
K_1 - a constant which represents frictional losses
K_2 - a constant representing aerodynamic drag

My question is why do we have both a V_g (ground speed) and V_a (Air speed)? Wouldn't these two always be the same?

LesterOfPuppets

I think the air speed might be there to take wind into account. I haven't looked at the article yet, however.

Shimagnolo

The frictional losses are caused by the rotating mechanical parts on the bike and the tires, so ground speed is the only thing that matters to it.

The aero drag is entirely dependent on the relative wind.

The only time V_g and V_a are the same is when the wind speed is zero.

LesterOfPuppets

The only time V_g and V_a are the same is when the apparent wind speed is zero.

prathmann

Don't understand what you're trying to say here. If I'm going 20 mph due south with a 20 mph wind out of the north then the apparent wind speed (i.e. what I feel) will be zero. But in that case V_g is 20 mph and V_a is 0 mph; certainly not the same.

LesterOfPuppets

Oops. Wasn't reading the variables properly there. You're correct.

jshort2010

Suppose V_a is the cyclists ground speed with respect to the wind. If we have a tail wind of 8 m/s (so 18 mph approx) and your going with the wind, then V_a = 0 if you are also moving 8m/s.

Thus P = g * m * 8 * (K_1 + s).

If we are going on level terrain than s = 1.

Assume the bike is 8 kg (so a pretty light one) and the rider 75 kg;

K_1 = 0.0053 approx, so K_1 + s = 1 approximately.

Thus P = g * m * 8 = g * 664.

g = 9.8 m / s^2

Thus P = 6500 watts approx.

This sounds wrong. The world record is around 2300 watts, and in the above case the 75 kg rider is going with the wind on a light 8 kg bike at only 8 m / s (18 mph).

This doesn't look right

Shimagnolo

On re-reading that formula, two things caught my attention:

- V_a must be *relative* wind, not wind speed. i.e. the apparent wind felt by the rider.

- That last V_g term makes no sense. I think that was an error made by the OP.

jshort2010

No I copied it correctly. Unless whoever wrote the wiki entry got it wrong.

jshort2010

I see where I made the mistake here. On level terrain s = 0 not 1.

Thus we have

P = 9.8 * 83 * 8 * 0.0053 = 34.5 watts.

That sounds a little low even with a 8 m / s ( ie 18 mph) tailwind.

Shimagnolo

Lots of formulas and calculators here: https://www.analyticcycling.com/

fietsbob

Force 10+ Gale as head wind, air speed can be 60 MPH ,
while you walk the bike because your ground speed is so difficult
you fall over if you try pedaling the bike.. ground speed zero, or backwards.

No formula needed, I live on the coast, and winter storms can arrive
with a vengeance.

Winter off SW Irish Coast, they are as strong or stronger
coming off the North Atlantic.

Air speed/ ground speed is an Airplane issue..

In a Boat , if the tide is coming in and you want to go out, right then,
and the water is moving against you.
probably better to drop the hook and wait a while.

LesterOfPuppets

Are there enough variables to account for running TT discs front and rear?

prathmann

Yes, the aerodynamic characteristics of the bicycle and rider are included in the K_2 constant which depends on the coefficient of drag of the bike/rider and the effective frontal area.

To Shimagnolo's comment on the last V_g:
The formula given is correct. The drag force due to air resistance is proportional to the wind speed squared - hence the (V_a)^2. But the power needed to move at a given speed is: P = V x F, so you multiply the drag force by the ground speed of the bike.