I'm interested in short, or trick, take-offs - such as from platforms, tall trees, etc.

I think that I should have

This will let me add to my intuition from regular launches (from sites with known-good launch conditions), and estimate how much velocity I need to add via a run / push.

The methodology to count the wind measure seem a bit more grey, right now.

I can have a sense of the wind where I am, but it may quickly change beyond my launch site. Since I'm considering how to launch from a stationary position, it would be good to feed my senses / calculations more data like this.

  • $\begingroup$ You can objectively measure the wind speed with wind speed meter (anemometer). $\endgroup$ Commented Dec 18, 2013 at 10:25
  • $\begingroup$ This is a rather open-ended question. It is possible to do a cliff launch where you simply step off the cliff, holding the glider pointing steeply downwards. In this case your required takeoff run is zero. I'm not sure whether or not this can be safely done in no wind-- perhaps off a platform projecting far in front of the cliff like a diving board? So a key question is, does your question intend the glider to be up to a normal flying airspeed before reaching the end of the launch run, or is it allowed to gain speed by falling a ways? $\endgroup$ Commented Jul 24, 2019 at 16:50

2 Answers 2


What you need is enough lift to remain in the air. For that, the maximum lift of your hang glider needs to exceed the total weight.

When you know the total weight of the glider (you included), you can derive the theoretical minimum airspeed from the lift formula:

$$L_{max}=\frac12{\rho}V^2C_{max}S$$ $$V_{min}=\sqrt{\frac{2W}{{\rho}C_{max}S}}$$


  • $L$ = lift
  • $\rho$ = air density
  • $V$ = air speed
  • $C_{max}$ = maximum lift coefficient
  • $S$ = wing surface
  • $W$ = weight

Now you have to know your maximum lift coefficient. Suppose this is 1.2, and for safety and initial manoeuvring, let's add a margin of 40% to the weight.

Let's say your mass is 85 kg, and the glider has a mass of 35 kg. The total weight would be 120 kg — that is about 1,177N weight, which for the sake of safety margin is increased to 1,650N.

I'll estimate the surface of the wing to be $16m^2$. Assume a normal day, the air density will be about $1.225kg/m^3$.

Filling in the numbers is the formula:

$$V_{min}=\sqrt{\frac{2\times1650}{1.225\times1.2\times16}}\approx 12 \textrm{ m/s}$$

Any wind speed you get you can subtract from this figure: what is left you have to achieve by pushing, running or diving.

  • 2
    $\begingroup$ @Qantas94Heavy Thanks for editing, it is a major improvement. $\endgroup$
    – DeltaLima
    Commented Dec 20, 2013 at 7:34
  • 1
    $\begingroup$ Hopefully it's even better now ;) $\endgroup$ Commented Apr 17, 2014 at 1:29

An airspeed indicator is probably the most critical instrument one can have in any type of aircraft they happen to be flying. It can save your life. Airspeed should never be "grey". While understanding the theory is important, far more so is the performance in the air, in other words, thoroughly testing AT ALTITUDE the stall speed and characteristics, and speaking with people experienced in your type. And, as previously suggested, build in a safety margin and adhere to in when getting close to the ground.


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