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My goal is to estimate the lift coefficient of a glider aircraft based on three properties:

  • Climbrate [m/s]
  • Turnrate [turns/minute]
  • Speed [m/s]

For this I'm using the lift equation:

L=Cl*0.5*r*V^2*A

Currently these properties are unknown:

  • L
  • Cl
  • A

For r (density) we'll take the default of 1.23 kg/m^3.

I'm toying with the idea to use the climbrate as a surrogate for the lift, but I can't seem to wrap my head around the relationship between the lift force (Newtons), and the climbrate (m/s).

A derivative of the climbrate could be the kinetic energy per mass unit (Ke/m==0.5V^2), but this does not bring me any further.

A possibility I've been exploring is to equalize the lift as the following equation by abstracting all unknowns into the units of lift.

climbrate+G == F*s/A*kg == Cl*0.5*r*V^2

I'm not sure though whether I'm justified to do so by laws of nature and mathematics.

Once this relationship is clear it's almost a trivial task to solve for the lift coefficient.

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2 Answers 2

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If G in $m/s^2$ is the acceleration due to gravity then you cannot add this to climbrate in $m/s$ because the units are different.

If the glider is climbing or falling at a constant rate (which could be zero) then the vertical forces on it are balanced. If climb rate is zero then there is no drag force vertically, so the lift force $L$ upwards is balanced by the weight of the aircraft $W=mg$ downwards : $$mg = \frac12 C_L \rho A v^2$$

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  • $\begingroup$ If I understand correctly L == mg when the climbrate is zero. As such L > mg when the climbrate is positive and vice versa. With your first point in mind, when using vertical acceleration instead of climbrate, would a*m/A == 0.5Cl*r*v^2 make sense dimensionality wise? $\endgroup$ Apr 13, 2020 at 14:21
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    $\begingroup$ Your equation makes sense dimensionally but the weight $W=mg$ is missing. If there is an upward acceleration $a$ then $ma=L-mg$. This assumes that there is no drag in the upward direction. If there is drag $D$ then $ma=L-mg-D$. Note that $D$ depends on vertical velocity whereas $L$ depends on horizontal velocity. $\endgroup$ Apr 13, 2020 at 15:55
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The lift is independent of the climb rate. The glider climbs because it flies in rising air. How fast that air rises or falls has no influence on lift. Lift is determined by the mass and gravitational acceleration, but not climb rate.

If wing area is unknown, you cannot compute lift coefficient from what you have. You can only find a relation between the lift coefficients of two different data points when speed or turn rate change (and all other parameters stay the same). Lift coefficient is inversely proportional to the square of airspeed and the inverse of the turn rate as explained here.

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  • $\begingroup$ Thanks for the reference! I always understood that an abundance of lift resulted in a gain of altitude, and that when entering an thermal, the angle of attack increased while maintaining the center of gravity and thus maintaining forward momentum with a higher climbrate. Can you indicate where my mental model went wrong? $\endgroup$ Apr 15, 2020 at 11:44
  • $\begingroup$ @CorstianBoerman: Yes: Excess lift is only needed to accelerate the aircraft upwards. Once it climbs with a steady climb speed, lift equals weight. For climbing you need excess energy which can be supplied by rising air or a running engine. Lift is only needed to keep the aircraft in the air, regardless of vertical speed. We've had a similar discussion over at Aviation SE. $\endgroup$ Apr 15, 2020 at 15:54

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