This answer is excellent:-

Is there any equation to bind velocity, thrust and power?

except that it is (in practice) unusable, since all the units are missing.

I'm guessing we use meters for length (prop diameter), and watts for power.

I'm not sure what units we use for mass (grams or kilograms?) - but either way - neither of those work (grams gives answers 10x too small, and kg gives answers 1000x too large).

  • I have a 10,000 watt motor
  • My propeller is 1.25meters diameter
  • At full pelt, using bathroom scales on the wall, it pushes me with 35kg force

Using 0.8 for the prop efficiency, and 0.9 for the motor efficiency, and 1.225 grams/l for normal air density, the referenced formula gives 367448 grams thrust - 10 times too much.

So yeah... what are all the units supposed to be?

  • 2
    $\begingroup$ How far away is the prop from you? Did you account for dispersion of that thrust, deflection off of your body curves, etc.? Without knowing exactly how you are measuring I would guess there is a lot of wasted wind that you aren't capturing. $\endgroup$ Nov 7, 2019 at 23:11
  • $\begingroup$ Thrust is force, and it should not be measured in units of mass (grams, kg). $\endgroup$ Nov 7, 2019 at 23:26
  • $\begingroup$ Scales do not measure mass, so 35kg is not expressed in the correct units. $\endgroup$ Nov 7, 2019 at 23:27
  • $\begingroup$ My motor is a Radne Raket Aero 120cc mounted on a nirvana colibri paramotor, so, I hung my bathrooms scales on a wall, stood in front of the scales wearing my motor on my back, and went full throttle. I had my arm outstretched against the scales, and the push from the propeller deflected the needle by approximately 35kg. $\endgroup$ Nov 8, 2019 at 2:04
  • $\begingroup$ A clever way of measuring thrust which can only be improved by mounting the scale on a pole in the open and repeat the test there, so no interference from walls will spoil the result. Still, I would expect only a small increase in thrust. The scale indicated 35 kg but actually measured 350 N. There is your missing zero. $\endgroup$ Nov 10, 2019 at 7:25

1 Answer 1


Any good formula (that is, the one without magic numbers baked in) will accept any units as long as they are used consistently.

If you use SI, then use every figure in SI, and you'll get the result in SI.

Force is measured in Newtons, which is different to the unit of mass (kg), specifically $N = kg⋅m/s^2$. The factor is the gravity acceleration $≈9.81\ m/s^2$ (which comes from the original definition of kg). So your measured thrust is about 343 N.

Because people are so used to treating force (weight) as mass, the 'force' units like kilogram-force (kgf) or pound-force (lbf) are commonly used, often implicitly. But this only leads to traps like you experienced and should never be used scientifically.

Density is $kg/m^3$, which is numerically equivalent to $g/l$ you used.

(In the context of propellers, one needs to be careful about the units of rotation. Strictly speaking, the SI unit of angular velocity is $rad/s$. But in practice coefficients are defined for 'n' in revolutions per second, $1/s$ (or even revolutions per minute, rpm, which is a non-standard unit). Fundamentally, radian is a dimensionless ratio, so it does not affect the resulting units (over 1/s) when used directly in formulas, and the error is not always obvious).

  • $\begingroup$ Brilliant! thanks for this - the "out by 10" problem had my wondering if it's got something to do with gravity, but I wasn't sure where I'd gone wrong... Newtons, of course, was my mistake. I notice it is a very common recurring problem in air theory that the units are always left out - the NASA web site has tons of amazing stuff, which cannot be used because of that problem. It might seem obvious to the authors what units to use, but there are SO MANY different units and things everywhere, omitting them is just condemning most of the audience to future pain when they make mistakes like me $\endgroup$ Nov 8, 2019 at 2:08
  • $\begingroup$ @AnonCoward For those with a background in physics, engineering, other sciences, etc. (which I would argue is a large chunk of the audience), raw formulas with no units is the norm. It means you can use whatever units you choose or is convenient for the given problem (as Zeus demonstrated). It necessitates understanding of the formula, not just computation. A double-edged sword for those still learning, for sure. $\endgroup$
    – Shamtam
    Nov 8, 2019 at 3:26
  • 1
    $\begingroup$ If you're not familiar with it, it is worth reading up on dimensional analysis. It's a very simple concept that is very powerful when working with any sort of formula. More detailed Wikipedia link here $\endgroup$
    – Shamtam
    Nov 8, 2019 at 3:28
  • $\begingroup$ Yes - exactly - that's what makes units so powerful - you can do math on the numbers, while also doing appropriate operations on the units, which gives you a "sanity check" at the end... but only if you know the right units to begin with of course. On that note - there IS still a units issue with that reference I gave - nothing is cancelling out the "watts" part of the result. $\endgroup$ Nov 11, 2019 at 8:22

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .