Recent content by martonhorvath

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?

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...

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

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...

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