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I am trying to calculate the optimal flight climb angle based off of some basic power calculations for an aircraft on steady climb with an angle of theta. Doing a balance of forces I obtained the thrust force, then multiplied by velocity to obtain power. The power was time independent, thus multiplying the term for power by time I obtained the energy, and then I try to optimize this by trying to find an angle for which energy is minimized, however it seems only pi/2 radians (90 degrees) gives a minimum on the graph of energy consumption which I obtained. Did I do something wrong? Is my approach wrong or is the question in itself flawed? I attach my calculations below as images, thank you! enter image description here enter image description here enter image description here

PS: This is not for homework, this is an interest of mine, I am on my second year studying aerospace engineering and am doing this (studying the efficiency/power equations of a UAV) as a way of getting 'ahead of the curve'. Any feedback on my method is appreciated :)

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    $\begingroup$ Without having read the whole derivation, do you realise that there is typically value in travelling horizontally during the climb. You usually want to minimise your energy consumption over the whole flight. If you climb straight up (e.g. by exchanging speed for height) you spent least energy during the climb phase, but you didn't get any closer to your destination and so you consume extra energy for the longer cruise flight segment. $\endgroup$
    – DeltaLima
    Commented Sep 28, 2022 at 16:10
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    $\begingroup$ And I think you'd better ask to merge the two profiles $\endgroup$
    – sophit
    Commented Sep 28, 2022 at 16:33
  • $\begingroup$ I know is possible to merge two profiles but I don't remember how... I suppose the question is correct here but I think it would be more efficient if you post the whole idea of your project in one single place instead of spreading it among several posts, only my thought $\endgroup$
    – sophit
    Commented Sep 28, 2022 at 17:03
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    $\begingroup$ I accidentally created two accounts; how do I merge them? $\endgroup$
    – Bianfable
    Commented Sep 28, 2022 at 17:22
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    $\begingroup$ please DO NOT post important content only as images. it is very unwelcoming to people using screenreaders and prevents them from participating. you can use mathjax here to format the equation properly $\endgroup$
    – Federico
    Commented Sep 29, 2022 at 6:34

4 Answers 4

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Did I do something wrong?

No, if you have no thrust limitation, your result of a vertical climb is correct. Rockets ascend that way initially while air is dense to minimize the energy expended.

On the other hand, the formulas you copied into your question assume a shallow climb since they apply small angle approximations. Also, thrust is not explicitly expressed as a function of speed and the acceleration needed to stay at the same IAS during climb is also neglected.

To answer more precisely, it would help to know what aspect of the climb needs to be optimized. Generally, you should express the force equilibrium in a climb as precisely as possible and then derive the equation according to the quantity to be optimized. Where the derivation becomes zero you will find the optimum.

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The optimal flight rate of climb angle can be derived from Vy.

Basically, at Vy, one flys the plane at optimal angle of attack and optimal thrust output.

As we have learned, lift is less than weight in a climb. Lift requirement is cosine climb angle × weight. The "excess power" can be used to climb, with optimal Angle of Attack providing the least induced drag.

Practically, optimal angle of climb can be tested by measuring rate of climb per unit time at various climb speeds (angles at full throttle). This testing should also yield Vx, the best angle of climb per horizontal distance covered to safely clear an obstacle.

Another option worth considering, if there are no altitude restrictions, terrain, or obstacles, is cruise climb, which can yield slightly faster arrival time to destination.

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    $\begingroup$ Ah I see, so I look for an optimal v*sin(theta) not an optimal theta, thanks! $\endgroup$
    – Pablo
    Commented Sep 28, 2022 at 17:22
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I may have missed something but your drag equation seems naive and so gives the wrong sort of speed-drag curve. You need to include the induced drag component.

Vx is CAS for best angle of climb[minimum distance], Vy is CAS for best rate of climb[minimum time] Both are for specific aircraft configurations, changing something like wing area, engine power, or the shape of the propeller will change both the speeds and the resulting angle and rate.

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The easiest way to get the speed for best rate of climb is to write down its equation first and then make the substitutions/simplifications.


Being $\gamma$ the climb angle, the (simplified) equilibrium equations on the horizontal and on the vertical direction are:

  1. $T=D+W\sin\gamma$
  2. $L=W\cos\gamma$

with the vertical velocity aka rate of climb defined as:

  1. $V_v=V\sin\gamma$

Under the hypothesis of small climb angle (less than some 15°), $\cos\gamma$ is practically 1. Getting $\sin\gamma$ from the first equation and $W$ from the second one and substituting them in the third equation, one gets:

  • $\gamma=\arcsin(\frac{T}{W} - \frac{1}{L/D})$
  • $V_v=V(\frac{T}{W}-\frac{1}{L/D})$

Now we can substitute in the latter the relevant expression for lift and drag and then equate to zero $\frac{\partial V_v}{\partial V}$ in order to get the best rate of climb.

Since you have a constant thrust, the derivation of $V_v$ gives:

$\frac{\partial V_v}{\partial V}=0=\frac{T}{W}-\frac{3\rho V^2C{_{D_0}}}{2W/S}+\frac{2KW}{\rho V^2S} \Rightarrow V@\text{max} V_v$

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  • $\begingroup$ As usual for the usual impolite downvoter: "Please consider adding a comment if you think this post can be improved" And: "Downvote answers that are incorrect or don't provide sufficient information" – $\endgroup$
    – sophit
    Commented Oct 7, 2022 at 4:44

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