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In a VTOL, I have calculated the power required for take off and cruise which was not difficult at all. Between take off and cruise, there is a phase where the VTOL has to transition by tilting the rotors from 0 to 90 deg. I am finding it hard to calculate the power required for this transition. I feel this is very important because it consumes more power than take off since it is creating lift (hasn't reached the minimum velocity for the lift requirement) for balancing weight and also needs to provide thrust for forward motion. Given to me are:

  1. Rotor parameters
    • number of rotors
    • Rotor diameter
    • rpm
    • number of blades
  2. Fixed wing parameters
    • span and chord (rectangular wing)
    • lift and drag coefficients

What are the factors considered to calculate the power required for transition? Please note: Make as many assumptions/approximation as you like. I just want a simple formula for a basic understanding.

  1. The VTOL is a tilt rotor type with ducted fans
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  • $\begingroup$ Are your rotors ducted or unducted? What kind of VTOL is this? Depending on the specific configuration, the transition and cruise phases would need to consider the disk loading like in a helicopter. If, however, the wing is large and the disk surface is negligible, then you can treat the rotors as vectored thrust sources and just use high school trigonometry to solve the transitory phase. $\endgroup$
    – AEhere supports Monica
    Feb 11, 2019 at 9:08
  • $\begingroup$ Ok let's say I assume my wing to be large and disk area negligible.....How do I calculate the power now?....Because I feel integration is involved from 0 deg to 90 deg. $\endgroup$
    – sai teja
    Feb 11, 2019 at 10:49
  • $\begingroup$ Another comment if I may. The maximum power requirement will be to produce its own weight in thrust plus enough thrust to hold altitude in a downdraft while in vertical flight. If you set your safety limit at say 20 feet per second, a drag calculation could be made for the plane being raised at this speed vertically based on area presented in vertical flight. You may find this is more than enough for transition to horizontal. $\endgroup$
    – Robert DiGiovanni
    May 12, 2019 at 0:01

2 Answers 2

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Like Hobbes said. For instance, after take-off, tilt the thrust 10° forward. The cosine is 0.98 so you only lose 2% of propulsion lift. The sine is 0.17, so forward thrust is 0.17 * total thrust.

The craft will speed up until aerodynamic drag = forward thrust:

$$ T = D = C_D \cdot \frac{1}{2} \cdot \rho \cdot V^2 \cdot S$$ $$ V^2 = \frac{2 \cdot T}{C_D \cdot \rho \cdot S}$$

Then tilt to 20° where you have sin(20) = 34% of the propulsive thrust pointing FWD and 94% pointing UP, etc.

Note that it is usual to express the drag equation relative to wing area S.

A simple real-time approximation in a spreadsheet would look like underneath. 9810 N thrust pointed back 10 deg results in sin10*9810 = 1703 N forward thrust, which takes about 300 seconds to speed up the craft to 97 m/s. Of course, because of the aerodynamic lift the craft climbs, changing AoA and creating induced drag etc. The simulation should be extended with reduction in rotor thrust proportional to aerodynamic lift. But I hope the gist of things is clear enough.

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  • $\begingroup$ But you should take into account the lift generated from the wings also right $\endgroup$
    – sai teja
    May 8, 2019 at 15:12
  • $\begingroup$ Yes you should, which you can calculate once you know the speed. The forward thrust accelerates the craft, causing both wing lift and drag to rise every second until T = D. $\endgroup$
    – Koyovis
    May 9, 2019 at 0:23
  • $\begingroup$ So I have studied the procedure you have followed.What if we just keep constant thrust and tilt the angle from 0 to 90. The problem what I am facing is that when we do that , we reach the desired stall speed at an angle of 6 deg and so what do I do to completely rotate it to 90 deg without creating extra lift. $\endgroup$
    – sai teja
    May 16, 2019 at 13:59
  • $\begingroup$ Reduce the thrust proportionally. $\endgroup$
    – Koyovis
    May 16, 2019 at 14:10
  • $\begingroup$ Then you would be decelerating in forward flight right. $\endgroup$
    – sai teja
    May 17, 2019 at 5:40
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The lift of the wing is proportional with the square of the aircraft speed, see the lift formula. That tells you how much vertical thrust you need at each speed.

The vertical thrust of the rotors depends on the cosine of the tilt angle, horizontal thrust depends on the sine.

The horizontal thrust leads to the speed you can reach at that tilt angle, but you need the total drag of the aircraft to calculate this. That's info you don't have, it seems.

Integrals give me headaches, so first I'd calculate these parameters for let's say 10-degree increments in tilt angle and plot these, perhaps a curve will become apparent in the plot.

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  • $\begingroup$ For the drag ,I have the drag coefficient and the area right...... and after finding out the parameters,how do i calculate the the total power required for transitioning.......How's the plot going to help me $\endgroup$
    – sai teja
    Apr 8, 2019 at 12:57
  • $\begingroup$ The plot will identify where the peak in power demand is, and you can either measure the height of that peak from the plot, or calculate from the speed indicated in the plot. You have lift and drag for the wing, but do you have drag info for the fuselage? $\endgroup$
    – Hobbes
    Apr 8, 2019 at 13:12
  • $\begingroup$ How do I calculate the drag when I do not know the velocity at that angle $\endgroup$
    – sai teja
    Apr 11, 2019 at 5:40

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