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I'm slightly confused regarding the formula quoted here:

How to calculate the amount of thrust of a autogyro during auto rotation?

When they say thrust is equal to the weight. Do they mean use earth weight directly i.e. 50Kg or am I supposed to calculate thrust as weight force, i.e. F = m * g?

If its the latter can I just use summary of component forces to determine vertical thrust, resultant from a horizontal axis of travel?

Slartibartfast
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In vertical autorotation, with a constant sink velocity, the lifting force produced by the rotor is exactly equal to the weight of the gyro, since any difference between weight and lift would cause an acceleration, and we're considering a constant sink velocity, with zero acceleration.

In a dive under autorotation, with a stable dive path, the force produced by the rotor is higher than the weight of the gyro, since the tip-path plane of the rotor is tilted back, so you have a rotor drag component that has to be added (vectorially) to the vertical component (equal to the weight) in order to calculate the total rotor force.

xxavier
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  • One of these days the fixed wing folks will get what you are talking about. They think lift is less than weight in a climb. (Taking the engine away, and saying "now try it", can be helpful). Vbg and Va are faster than Vmin sink for reasons listed above. – Robert DiGiovanni Aug 19 '20 at 09:10
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    @RobertDiGiovanni Rather than perpetuating your viewpoint in comments, bypassing SE's quality control measures (the infamous Downvote...), perhaps it is more constructive to post your point of view on the appropriate question concerning fixed-wing flight. I'm quite eager (if not very sceptical) to see your arguments why you think lift would not be less than weight in a climb. – Sanchises Aug 19 '20 at 10:13
  • What I'm trying to get a hold on is for the formula: $$ T = 2v^2· A· rho $$ do I use the 'weight force' as the value for thrust? or do I calculate the thrust in newtons with: $$ T = F = m * a $$? – Slartibartfast Aug 19 '20 at 11:25
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    Thrust is a force, and the SI unit for force is the newton. If you multiply velocity squared times area times density, and you use m/s, m2 and kg/m3, you obtain the force in newtons. Automatically... Advantages of the SI system... – xxavier Aug 19 '20 at 12:49
  • You should learn to use simple dimensional analysis. You can find it explained in any elementary physics book... – xxavier Aug 19 '20 at 12:54
  • @Sanchises because the vertical lift is cosine lift, there for lift is greater than weight. Don't be fooled by "vertical thrust component". Thrust maintains airspeed, wing vertical lift renders plane "weightless". It works best that way. With the rotor, vertical drag spins the rotor, with the glider, vertical drag pushes it forward. Once moving, the wing lift keeps it steady state. Drag producing thrust, ha! (A windmill produces electricity, or could be a propeller with an electrical source). – Robert DiGiovanni Aug 19 '20 at 15:17
  • @Robert I agree that "thrust" is a poor choice of words for the force of a autogyro rotor - but then so is drag. Call it lift. (1/2) – Sanchises Aug 19 '20 at 18:34
  • @Robert Vertical force balance is $W=L\cos(\alpha)+T\sin(\alpha)$. Horizontal balance is $L\sin(\alpha)=T\cos(\alpha)$ (with $W$ weight, $L$ lift, $T$ net thrust, so thrust minus drag). Rewriting we get $$L=W\dfrac{1}{\cos(\alpha)+\dfrac{\sin^2(\alpha)}{\cos(\alpha)}} = W\cos(\alpha)$$ But once more, please post your point of view on the question linked. – Sanchises Aug 19 '20 at 19:26
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Weight is a force by definition. Its SI unit is newton. In most circumstances, weight is equal to $F=mg$.

The quantity in kilograms is called mass. In colloquial speech it is often called weight because we measure mass using the weight it exerts on a scale. In a technical context however, weight is always a force.

To make matters more complicated, there are units called "kgf" and "lbf"; kilograms-force and pounds-force respectively. One kgf is equal to the weight (force!) of one kilogram in Earth gravity.

Sanchises
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  • Fortunately, the 'kilogram-force' or 'kilopond' is forgotten and no longer used. Its use, together with the 'real kg', of mass, has caused untold mistakes, and not only by students... It could be used more safely if mass was accounted in 'Technical Mass Units' or TME... But it was all a mess... – xxavier Aug 19 '20 at 08:47
  • @xxavier If only the kgf was truly forgotten. It remains very appealing when talking about thrust or lift to many. I've lowered my expectations and as long as they use 'kgf' and not 'kg' when talking about forces (and refrain from using freedom units) I'm happy. – Sanchises Aug 19 '20 at 10:03
  • But for the formula in question they are talking about thrust being equal to Kilogram-Force as measured on a standard body scale for instance... yes? – Slartibartfast Aug 19 '20 at 10:31
  • Isn't force supposed to be measured in newtons, and mass in kilograms? – Slartibartfast Aug 19 '20 at 10:40
  • @nobody_loves_you Exactly. So 50kgf is about 490N. – Sanchises Aug 19 '20 at 11:28
  • @nobody_loves_you And the rules for which numerical value to use: the same rules apply as always when using mixed units. Ideally you convert everything to SI units before calculation, or your result will simply be in some nonsensical unit like foot-pounds per acre-ounce. – Sanchises Aug 19 '20 at 11:40