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I'm calculating the lift of a Cessna 172 and I don't understand the numbers given for weight. The maximum landing weight is 2550 pounds which is nearly 1134kg. But does that mean 1134 Newtons (weight) or 1134 kilograms (mass)?

Since weight = mass x gravity, I think the weight of the C172 is in Newtons even though it's written using a unit of mass. Am I correct?

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    $\begingroup$ No, but the person specifying weight in pounds could have been more explicit. It is pound-force (as opposed to pounds as a unit of mass), and you convert 1 Newton to 0.2248 pound-force. Somehow the insights of Isaac Newton or Galileo Galilei never found their way into the English language or the imperial system. Use the metric system - it is better. $\endgroup$ – Peter Kämpf Dec 4 '15 at 7:12
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    $\begingroup$ @PeterKämpf *mutter* The English language copes just fine with the concepts of mass and weight, and the system of units you're referring to is American customary units, not Imperial (the two agree on the pound but have different tons and different liquid measures). $\endgroup$ – David Richerby Dec 4 '15 at 9:12
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    $\begingroup$ As I understand your question, I believe you to be technically correct. However, remember that U.S. aircraft manufacturers typically publish weight limitations in both pounds and kilograms even though the first is a unit of force and the latter a unit of mass, and use a simple constant to convert between the two. Boeing uses 0.45359237 pounds per kilogram. This is a simplification that is workable because the variation in gravity between the surface of the earth the max altitude of aircraft is negligible for the practical purposes of aircraft weight limitations. $\endgroup$ – Terry Dec 4 '15 at 9:27
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    $\begingroup$ The question has an obvious answer: the mass of the C172 is about 2500 lbs, 1130 kg. This is what is sometime referred as weight in layman terms. If someone asks you your weight, usually you provide your mass for the answer and you don't use Newtons. In aviation MTOW is actually the maximum mass. $\endgroup$ – mins Dec 4 '15 at 12:01
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    $\begingroup$ @mins, So you mean the weight is 1130 X 9.8 kgm/s^2 ? $\endgroup$ – wonwoo Dec 4 '15 at 14:19
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The maximum landing weight given here is actually the mass, not weight.

In general, the various 'weights' given for aircraft are masses, not weights. i.e. they are in kilograms, not Newtons.

This is simply an extension of our everyday usage. As @mins pointed out, if someone asks your weight, you tell your mass (70 kg or whatever), not the weight itself (~686N).

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    $\begingroup$ This is correct - since Gravity is (near enough) constant on/near Earth (~9.81 N/kg), mass can be used in lieu of weight - as long as you specify units to avoid ambiguity, anyway. When non-scientists talk about weight they almost always mean mass. $\endgroup$ – Jon Story Dec 4 '15 at 15:42
  • $\begingroup$ Actually, the actual limit is weight, because it is about the induced drag needed to generate corresponding amount of lift and the thrust needed to overcome that and those are all forces. However, there is a general habit of quoting mass for weight and on Earth it works fine. $\endgroup$ – Jan Hudec Jun 6 '16 at 20:48
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On Planet Earth, 1kg of mass exerts 9.81 Newtons of force (weight). Because this is true anywhere on planet Earth to within a tiny fraction of a percent, we use kg as a convenient measure of weight, even though pedantically it is a measure of mass.

If you are thinking of landing your C-172 on the Moon, Mars, or some other locale where G is different from 9.81 ms-2, then you'll have to convert kg to Newtons of weight. Otherwise, you should consider kg a unit of both mass and weight.

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  • $\begingroup$ Pretty large fractions of a percent, actually. Measured accelerations of gravity on Earth's surface vary between 9.76 and 9.83 N/kg, according to Wikipedia. (Also, the symbol for kilogram is kg, not KG). $\endgroup$ – hmakholm left over Monica Dec 4 '15 at 17:32
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    $\begingroup$ 9.76 is only found on 22,000 ft. mountains near the equator. And even that differs from 1G (9.81) by approx 0.5%. $\endgroup$ – abelenky Dec 4 '15 at 17:58
  • $\begingroup$ Landing a C172 on the Moon, or Mars will be difficult, as flying requires lift which requires a fluid to fly into. $\endgroup$ – mins Dec 4 '15 at 18:57
  • $\begingroup$ You think lack of a fluid is the most significant problem there?? I think you've got several dozen insurmountable obstacles before fluid-dynamics even enters into the picture/ $\endgroup$ – abelenky Dec 4 '15 at 18:59
  • $\begingroup$ "One half" is not a "tiny fraction" in my world ... $\endgroup$ – hmakholm left over Monica Dec 4 '15 at 20:01
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Reiterating the point made by @aeroalias, the "weight" that FAA/manufacturer talks about is really "mass" in the pure physics world. However, in the real world, imagine you have a big enough scale to hang the airplane to. The MTOW or Max Take Off Weight would be the reading on that scale. So, if you are weighing your passengers' baggage, use a reliable scale. You can then use it directly in your weight and balance calculations.

So, don't worry about conversions. Just use the real world "weight" in Kgs or Lbs as the case may be as long you are flying on earth.

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  • $\begingroup$ The FAA weight is the force of the mass on Earth. Divide the weight by 32 to get the mass in SLUGS. $\endgroup$ – user3344003 Dec 4 '15 at 17:00
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Aircraft are weighed on scales. Attempting to measure the mass of a loaded aircraft isn't practical. Clearly, the figure indicates weight.

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  • $\begingroup$ The other answers state the opposite. Would you happen to know something the others don't? $\endgroup$ – Federico Dec 15 '17 at 6:34
  • $\begingroup$ @Federico, I'm not sure whether you are referring to my two initial postulates or my conclusion. While the plane's mass and acceleration ultimately determine the landing load, in practical applications, the starting figure for calculating landing weight is one directly measured by scales placed under the landing gear. $\endgroup$ – Jim Perris Dec 22 '17 at 21:28
  • $\begingroup$ Everyone says that the figure indicates weight, you are the only one claiming it does not. Moreover, even if direct measure is impractical, there are well established estimation methods. $\endgroup$ – Federico Dec 22 '17 at 21:55

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