I have a problem that it gives us the MTOW of a UAV (200 kg), the speed and the reference area. So its probably easy to find the CL knowing that it flights at steady level flight (L = W). I'm confused about the use of MTOW. Should we multiply it by the g? (gravity acceleration)

I've done some exercises in the past using lbs as the unit of MTOW and we didnt multiply it with the g. (John Anderson. Aicraft performance and design) Is it different when using lbs and kgs? I'm confused with the different metric/imperial systems.

  • 1
    $\begingroup$ I feel you are confused about basic physics (difference between mass and force) rather than SI vs. Imperial units. If your question asks for the CL while the craft flies at that MTOW (a simple approximation), and you can assume some standard air density, then this is an elementary algebra question where you only need to solve the lift equation to obtain the CL. Or it may be a poorly written question that needs you to make some extra assumptions to solve it. $\endgroup$ May 11, 2020 at 12:08
  • $\begingroup$ I do understand what you are saying. But I'm a little confused about the use of MTOW. I've seen Anderson uses the equation CL = (2*W) / rhoSV^2 and the W is at lbs. In my case I've got a MTOW of 200kg. Can I use it as 200kg to calculate the CL? $\endgroup$
    – ChrisP
    May 11, 2020 at 12:16

2 Answers 2


For every equation the left and right side should have the the same units. For example:

$$ Speed = \frac{distance \:[m]}{time \: [s]}$$

Or in units:

$$[\frac{m}{s}] = \frac{[m]}{[s]} $$

The same so with the lift equation:

$$L = C_L \cdot \frac{1}{2} \rho V^2 S$$

With units:

$$L [N] = C_L [-] \cdot \frac{1}{2} \: \rho \: \left[ \frac{kg}{m^3} \right] V\left[\frac{m}{s} \right]^2 S [m²]$$

Units only: $$\require{cancel} [N] = [-] \cdot \left[ \frac{kg}{m^3} \right] \: \left[\frac{m^2}{s^2} \right][ {m^2}] $$ Simplifying: $$\require{cancel} [N] = \left[ \frac{kg}{{\cancel{m^3}}} \right] \: \left[\frac{m^\cancel{2}}{s^2} \right][\cancel{ {m^2}}] = \frac{[kg][m]}{[s^2]}$$

Which is correct, link to Wikipedia.

So if you want to use the equation:

$$ L = W $$

Both have to be in $[N]$. You have two options, either have it given to you in Newton, or convert a weight (in $[kg]$) to $[N]$ by using the gravitational acceleration g (in unit $[\frac{m}{s^2}]$ )

I think that the confusion stems from the fact that the unit pound is used in two ways.

  1. for weight called pound (mass) $$1 [lb] = 0.45359237 [kg]$$
  2. for force called pound (force)
    $$1 [\mathbf{lbf}] = 1 [lb] \cdot g = 4.44 [N] $$

So if somebody is unspecific and asks what the lift is needed to carry 400 pounds, we don't know if it's way 1. or 2.


For any kind of applied math that you do, the units have to be self-consistent.

For example, we know that gravitational force is:


If you want $W$ in lb [force], and provided that $g$ is given in 32.2 ft/s2, then $m$ has to be provided in slug. If you use kg, then you have to first convert it to slug.

Units can also cancel. Units cancel out when the unit is divided by the same unit.

For example, $\frac{1ft}{1ft}=1$; but if I have in on the denominator, then $\frac{1ft}{1in}=12$, because there are 12 inches in 1 ft.

Back to the lift coefficient, which is unitless. Assuming lift is equal to weight, then

$$C_L=\frac{2W}{\rho V S^2}$$

If $W$ is in lb, then a self consistent set of units would be: $[\rho]=slug/ft^3$, $[V]=ft/s$, $[S]=ft^2$, noting that the definition of $lb=slug*ft/s^2$.

If you have a mass in kg, then an obvious derivation of weight would be to Newton, which is kg*m/s2. So you may have: $[W]=kg*m/s^2$, $[V]=m/s$, $[\rho]=kg/m^3$, $[S]=m^2$.


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