# How can the required wing area for climb be calculated, given aspect ratio? I'm currently working on problem 3.20 from the picture above. In order to find the rate of climb, I'm using the formula Rate of Climb= (Power available - power required)/weight. Doing this, I need to find the power required, which the formula I'm using is Pr=sqrt((2*W^3*Cd^2)/(densitywing areaCl^2)). Now I have everything but the wing area needed for this climb. Given the aspect ratio of 12 and a Cd of .01, how do I calculate this. Also, please let me know if there is a simpler way of going about this problem. Thanks

• You get the wing area from solving 3.19. – Peter Kämpf Nov 2 '16 at 23:57
• I didn't realize, but from problem 3.19 what type of propulsion system should I use for calculations? – user17495 Nov 3 '16 at 2:13
• It's a human powered aircraft. The text even mentions the Gossamer Condor. Isn't the propulsion system obvious? – Peter Kämpf Nov 3 '16 at 6:13
• That was my mistake. But I am still having a hard time finding the wing area from problem 3.19 because the cruising flight velocity formula involves wing area and required wing area for climb requires a velocity? What formula can I use with the given variables? – user17495 Nov 3 '16 at 20:14
• Could you edit question and add references of that book ? Might be of some interest for many people here. – kebs Nov 4 '16 at 22:15

## 2 Answers

Let's start with 3.19, since you need the result for 3.20. The parameters given suggest that we should calculate drag $D$ with the parabolic formula $$D = \frac{\rho}{2}\cdot v^2\cdot S\cdot\left(c_{D0} + \frac{c_L^2}{\pi\cdot AR\cdot e} \right)$$ I haven't seen the symbol $R_A$ before, but its magnitude suggests it is the aspect ratio (normally $AR$ in the US, $\Lambda$ elsewhere). Since the whole is an optimisation problem (we look for flight with minimum power consumption), we can determine the lift coefficient $c_L$ already. For propeller aircraft the point of minimum power is when the lift coefficient is $$c_L = \sqrt{3 \cdot \pi \cdot AR \cdot e\cdot c_{D0}}$$ Hey, you know all terms already! This allows you to proceed to the wing area once the needed lift is known. My next heroic assumption is that $W$ actually denotes a mass (the unit seems to say it is pound-force). The power plant should weigh 100 - 150 lbs already, so 200 lbs for the whole aircraft seems reasonable. So you can proceed: $$L = 200 lbs \cdot g = \frac{\rho}{2}\cdot v^2\cdot S\cdot c_L$$ Plug in what we know, use sea level density ($\rho = 1.225 \frac{kg}{m^3}$) and the previously calculated lift coefficient: $$S\cdot v^2 = 1527 \frac{m^4}{s^2}$$ This gives you pairs of wing area and speed squared which together fulfil the requirement. Some pairs, though, make more sense than others. Generally, the biggest wing will need the lowest power but will become increasingly fragile as wing area increases.

But there is still one requirement left: We cannot use more power than those 0.33HP: $$P = 0.33\cdot0.7 HP = 172 W = D\cdot v = S\cdot v^3\cdot\frac{\rho}{2}\cdot0.04$$ $$\rightarrow S\cdot v^3 = 7094.8 \frac{m^5}{s^3}$$

Both conditions can be fulfilled when the flight speed $v$ is 4.646 m/s. This makes the wing area $S$ = 70.735 m².

Solving this for Denver, CO and finding the climb speed with 0.6 HP of power should be easy now. Just calculate the SEP and you know the climb speed.

I believe what you are missing here is the fact that this kind of an aircraft would cruse at best $L/D$ ratio.

1. Solve for $CL$ which gives best $L/D$ from the given $Cd_0$ and $Cl_0$ (parabolic drag model)
2. with this $CL$ solve for velocity and wing area.
3. Power required $(D*V)$ and Rate of Climb can also be found from the same drag model.

Hope this helps.

• Best L/D is the point of lowest drag. I would prefer to fly at the point of lowest power consumption. – Peter Kämpf Nov 4 '16 at 7:25