Optimizing energy expenditure over a race course in cycling

[martonhorvath]

Homework Statement

I created an algorithm for calculating the time gain resulting from "extra" power output over headwind or tailwind sections to model different pacing strategies in cycling and cross-country skiing. A course is modelled as an uphill and downhill consisting of the same distance with the same incline ($\alpha$).

As a next step, I want to assume a "pool" of W as a capacity for the given course (same distance against headwind and tailwind(s)) and calculate the optimal distribution of work and power output over the section for maximising average speed over the course (2s).

Relevant Equations

$W_{total} = W_{1} + W{2} = W_{total}×k + W_{total}*(1-k)$, an optimal $k$ is needed to be found for maximal average speed over the course (headwind and tailwind together).

$$W = W_{gravity} + W_{friction} + W_{air}$$

Dividing by s:

$$F_{total} = mg(sin(\alpha)+\mu cos(\alpha))+0.5×C_{d}A\rho×(v+v_{wind}×sin(\beta))^2$$

Then expressing v for both sections separately:

headwind:

$$0.5×C_{d}A\rho×v^2+v_{wind}×sin(\beta)×C_{d}A\rho×v+mg(sin(\alpha)+\mu cos(\alpha))+0.5×C_{d}A\rho×(v_{wind}×sin(\beta))^2-F_{1}=0$$

tailwind:

$$0.5×C_{d}A\rho×v^2-v_{wind}×sin(\beta)×C_{d}A\rho×v+mg(sin(\alpha)+\mu cos(\alpha))+0.5×C_{d}A\rho×(v_{wind}×sin(\beta))^2-F_{2}=0$$

where $F_{1}=F_{total}×k$ and $F_{2}=F_{total}×(1-k)$

Then I used the quadratic formula but got clearly wrong results for $v$. If $v$ would be correct then I would continue with calculating the times spent in each section, then average speed.

I would like to ask for some help because I don't know where I mistake.

 

[haruspex]

I assume beta is the wind angle. You have written the expression for the total drag force, but you only want the component opposing v.
What is this $mg\mu\cos(\alpha)$ term? Friction doesn't act against the cyclist unless braking. It is static friction that allows the cyclist to progress. There is rolling resistance. Is that what it represents?
In expanding the $(v+v_{wind})^2$ term you forgot to double the $vv_{wind}$ term.

In what way was your result clearly wrong?

 

[martonhorvath]

I assume beta is the wind angle. You have written the expression for the total drag force, but you only want the component opposing v.
What is this $mg\mu\cos(\alpha)$ term? Friction doesn't act against the cyclist unless braking. It is static friction that allows the cyclist to progress. There is rolling resistance. Is that what it represents?
In expanding the $(v+v_{wind})^2$ term you forgot to double the $vv_{wind}$ term.

Correct, beta is the wind angle, but I don't see why it is not written for the opposing component, for the first assumption, the one I could solve, it worked. The truth is that I mainly made this for cross-country skiing, where friction is very significant compared to cycling meaning a $\mu=0.018$, but it can be the rolling resistance as well. Thanks, that was a typing mistake, I worked on this in Excel, and it was correct there.

 

[martonhorvath]

The result is wrong in a way that the speed value I got is clearly too low.

 

[haruspex]

Correct, beta is the wind angle, but I don't see why it is not written for the opposing component

Actually, I misdescribed your error.
The relative velocity has head-on component $v+v_w\sin(\beta)$, and a lateral velocity $v_w\cos(\beta)$. The magnitude is therefore $\sqrt{v^2+v_w^2+2vv_w\sin(\beta)}$ and the total drag force is proportional to the square of that.
The component of that opposing motion is $(v+v_w\sin(\beta))\sqrt{v^2+v_w^2+2vv_w\sin(\beta)}$.

You are taking $\beta=0$ to mean a crosswind, yes?

The result is wrong in a way that the speed value I got is clearly too low.

I have no way to judge because you have not stated any values for most of your variables or your result.

 

[martonhorvath]

The result is wrong in a way that the speed value I got is clearly too low.

Actually, I misdescribed your error.
The relative velocity has head-on component $v+v_w\sin(\beta)$, and a lateral velocity $v_w\cos(\beta)$. The magnitude is therefore $\sqrt{v^2+v_w^2+2vv_w\sin(\beta)}$ and the total drag force is proportional to the square of that.
The component of that opposing motion is $(v+v_w\sin(\beta))\sqrt{v^2+v_w^2+2vv_w\sin(\beta)}$.


I have no way to judge because you have not stated any values for most of your variables or your result.

Thank you for the formula, I have also thought that using only $v_{wind}×sin(\beta)$ is an oversimplification.

Yes I take $\beta=0$ for corss-wind.

It has turned out that my calculations are not as wrong as I thought, just made a stupid and fatal mistake when compared to the other method (i.e., I compared not the same situations) but a marginal difference, usually a few hundredths or tenths remained still. I will try to find the cause of that then velocity calculation should be fine, therefore section time and average speed over the course as well.
Do you have an idea how to find the optimal $k$ for maximal $v_{ave}$?

 

[haruspex]

Thank you for the formula

Do you see how I derived it?

Do you have an idea how to find the optimal k for maximal vave?

You want to minimise the total time, so write an expression for that and differentiate wrt k.

 

[martonhorvath]

Do you see how I derived it?

Calculating the resulting velocity: $\sqrt{(v+v_{w}×sin(\beta))^2+(v_{w}×cos(\beta))^2}=\sqrt{v^2+2vv_{w}×sin(\beta)+(v_{w}×sin(\beta))^2+(v_{w}×cos(\beta))^2}=\sqrt{v^2+v_{w}^2+2vv_{w}×sin(\beta)}$ I can see it until here, but how does the multiplication by $(v+v_{w}×sin(\beta))$ come?

 

[haruspex]

Calculating the resulting velocity: $\sqrt{(v+v_{w}×sin(\beta))^2+(v_{w}×cos(\beta))^2}=\sqrt{v^2+2vv_{w}×sin(\beta)+(v_{w}×sin(\beta))^2+(v_{w}×cos(\beta))^2}=\sqrt{v^2+v_{w}^2+2vv_{w}×sin(\beta)}$ I can see it until here, but how does the multiplication by $(v+v_{w}×sin(\beta))$ come?

If you have a vector magnitude m in the direction (x,y) then its component in the x direction is $m\frac x{\sqrt {x^2+y^2}}$.
In the present case, m, the drag force is $c(x^2+y^2)$, so we get $c x{\sqrt {x^2+y^2}}$.