Physics & Physiology

....wow.....this edit got f'd up......

Cool stuff! Not really sure how to apply it in a plactical application but I suppose I might be four times a likely to [censored] in my bibs going 60 as opposed to 30.😄

jhuskey said:

Cool stuff! Not really sure how to apply it in a plactical application but I suppose I might be four times a likely to [censored] in my bibs going 60 as opposed to 30.😄

Hey, that's a good insight! I didn't think of "physiological impact" that way. I think, now we've got all the dynamical variables covered: forces, energy, and excrement. For me, excremental impact occurs when I enter a corner that can be safely taken at velocity, v, while moving at a velocity, 3v.

9606 said:

Is my general rule of thumb correct (for a cyclist):

Frontal & surface friction increase by the square when velocity doubles.
Power requirements increase by the power of three when velocity doubles.
The physiological impact increases by the power of four when velocity doubles.

Aero drag forces increase with the square of the velocity. Power requirements increase as the cube of the velocity. If the velocity increase 3x and then 5x, then drag increases 9x and 25x, while power needs increase 27x, and 125x.

I don't know what you mean by "physiological impact." There're really aren't any higher order velocity dependent parameters.

NOTE: This should have been above where the f'd up edit is. When doing the edit I had an increase in entropy that behaved as a Heaviside function.

Found an article that talks of the "physiological impact", and answers my question.
Intensity Factor and Training Stress Score | Two Wheel Blogs

9606 said:

Is my general rule of thumb correct (for a cyclist):

Frontal & surface friction increase by the square when velocity doubles.
Power requirements increase by the power of three when velocity doubles.
The physiological impact increases by the power of four when velocity doubles.

The powers are correct, but the "when velocity doubles" part is confusing.

When velocity doubles, drag force increases by 2^2, power required by 2^3, and physiological stress by 2^4.

If velocity increased by 10%, then drag increases by 1.1^2, power by 1.1^3 and phys. stress by 1.1^4.

frenchyge said:

The powers are correct, but the "when velocity doubles" part is confusing.

When velocity doubles, drag force increases by 2^2, power required by 2^3, and physiological stress by 2^4.

If velocity increased by 10%, then drag increases by 1.1^2, power by 1.1^3 and phys. stress by 1.1^4.

What if velocity increases by 8i % ?

Clearly drag decreases by .64-16i %, power decreases by 1.9+24i% and phys. stress decreases by 3.8+32i%.

All the more reason to get on the bike instead of imagining yourself going faster, since imaginary velocity increases clearly reduce power, although they are good for relieving stress.

Edit: Upon further thought, I realized that of course, if you increase velocity by roughly 57.7i%, you'll actually have no power requirements at all, assuming your initial velocity was real.

alienator said:

Aero drag forces increase with the square of the velocity. Power requirements increase as the cube of the velocity. If the velocity increase 3x and then 5x, then drag increases 9x and 25x, while power needs increase 27x, and 125x.

So what other resistive force increases at greater rate than aerodynamic drag to drive the rate of increase in power required?

jollyrogers said:

So what other resistive force increases at greater rate than aerodynamic drag to drive the rate of increase in power required?

Aero drag--form drag, skin friction-- is the only one with a velocity squared dependence. Rolling resistance is linear with velocity. Bearing drag is linear with velocity. Gravitational force is linear with acceleration and mass. Seal drag is linear.

alienator said:

Aero drag--form drag, skin friction-- is the only one with a velocity squared dependence. Rolling resistance is linear with velocity. Bearing drag is linear with velocity. Gravitational force is linear with acceleration and mass. Seal drag is linear.

If power required increases cubically with velocity than resistive forces must increase cubically with velocity. Something isn't right with these rules of thumb.

Speaking of drag, whatever happened to that cross-dressing dude that use to post here?

jollyrogers said:

If power required increases cubically with velocity than resistive forces must increase cubically with velocity. Something isn't right with these rules of thumb.

No, you're not looking at quite right. If you look at the dimensions of force and power you, you see that force goes as length (m/(s^2), while power goes as length squared ((m^2)/(s^3)), so if drag force varies with velocity squared ((m^2)/(s^2)), then power to overcome drag force has to vary with velocity cubed ((m^3)/(s^3)).

It's even easier if you look at the equations for each and do the dimensional analysis. When done that way, you find that the units of power are correct when velocity cubed is used.

I've never liked dimensional analysis all that much. I'd rather just say that power is force times velocity, and force ~ a*v^2, velocity = v, so power = a*v^3 where a is a constant.

9606 said:

The physiological impact increases by the power of four when velocity doubles.

The physiological impact increases (roughly) by a factor of sixteen when power doubles.

acoggan said:

The physiological impact increases (roughly) by a factor of sixteen when power doubles.

So,

PL=KxP^3
P=KxV^3

PI=K x V^6

doubling velocity leads to x64 the physiological load.

So if i can ride for 64 minutes at 35 kph, maybe I can ride at 70 for one minute. I can only hope.

11ring said:

So,

PL=KxP^3
P=KxV^3

PI=K x V^6

doubling velocity leads to x64 the physiological load.

So if i can ride for 64 minutes at 35 kph, maybe I can ride at 70 for one minute. I can only hope.

Whoops, should be

PL=KxP^4
P=KxV^3

PI=K x V^7

doubling velocity leads to x128 the physiological load !

Lucky i can still ride at 35 kph for 128 minutes !

Enriss said:

I've never liked dimensional analysis all that much. I'd rather just say that power is force times velocity, and force ~ a*v^2, velocity = v, so power = a*v^3 where a is a constant.

Nothing wrong with dimensional analysis, it can simplify a lot of difficult concepts. But I like the way you break it down as it helps when considering windy conditions.

Neglecting acceleration:
Power = V_ground * (sum of all resistive forces)

One of those resistive forces is air resistance and:
F_air-resistance is proportional to V_fluid^2
where V_fluid is speed at which the cyclist moves through the air or the vector difference of wind velocity and ground velocity (or sum if you choose your reference directions appropriately).
So in calm conditions power due to air resistance is proportional to V_ground^3
but in windy conditions it's really:
P_air-resistance is proportional to V_ground*(V_ground-V_wind)^2

If you consider riding at 10 m/s (~ 22 mph), with a 3 m/s (~ 7mph) tailwind you'd get:
P_air resistance is proportional to 10*(10-3)^2 =490
which is different than some folks expect at (10-3)^3 or 343

The attached equations describe things nicely and more completely in two dimensions.

So your approach of breaking out power into velocity and net resistive force and then defining one of those resistive forces as being proportional to V^2 is useful in real world situations.

-Dave

Dimensional analysis is next to godliness.