In a recent question on the Aviation Stack Exchange, one explained that would cause an autogyro's crash. Basically, it says that the one thing that would make the gyro crash is to unload the motor too much, as explained here:

[...] Unfortunately, as soon as the rotor stops spinning, the whole aircraft falls like a brick and the rotor may be impossible to restart in flight. This is a situation that should be avoided at all costs.

Normally it is not a problem since the weight of the aircraft keeps the rotor spinning. However, if the weight becomes too low or even negative, the angle of attack will become negative, and the rotor will slow down and eventually stop. It can happen when the pilot "pushes on the stick" and dives.

That's this bold text that caught my attention. I learned in high school that the weight of an object equals its mass in kg times the acceleration of gravity in N/kg. An object mass cannot be negative and AFAIK, neither does the acceleration of gravity. So how can it be possible that the weight is negative?

  • $\begingroup$ That mass x g formula is for an object under no vertical acceleration. $\endgroup$ – Mike Dunlavey Mar 30 '18 at 13:26
  • $\begingroup$ Why didn't you ask the person who posted the answer in Aviation SE? $\endgroup$ – sammy gerbil Mar 30 '18 at 22:19
  • $\begingroup$ @sammygerbil Because the quote in my question is not his, but those of a reference given in his question. He asked about engine mechanics, nothing indicated he was skilled in physics (plus it would have been off-topic to talk about this on Aviation.SE). $\endgroup$ – avazula Mar 31 '18 at 11:59
  • $\begingroup$ If someone quotes from a reference to explain his answer, it is reasonable to assume that he understands what the quote means. This answer has 130 votes : the Aviation community has judged that Wayne Conrad knows what he is talking about. Negative weight is a term used in aviation, so a person using Aviation SE who used the term in his answer would be expected to know what it means. The explanation of aviation terminology is not off topic in Aviation SE. Negative weight is not a term used in physics. Physicists can only guess what aviators mean by it, unless they are also aviators. $\endgroup$ – sammy gerbil Mar 31 '18 at 19:05
  • $\begingroup$ I'm sorry, I'm an engineer in embedded systems, I didn't know that there's no such thing as negative weight in physics. I knew the notion of weight and g and couldn't find the answer on internet, so I asked. $\endgroup$ – avazula Apr 1 '18 at 21:09

The weight of an object is the force given by Newton's second law:

$$ F = ma $$

As you say the mass is constant, and if an object is neither accelerating up nor down the acceleration is just the gravitational acceleration $g \approx 9.81 \text{ms}^{-2}$. Then we get the familiar equation for the weight:

$$ W = mg $$

However suppose you are doing a loop. If you've ever done this you'll remember that your weight increases at the bottom of the loop and decreases at the top of the loop. That's because at the bottom of the loop the plane is accelerating upwards and at the top of the loop the plane is accelerating downwards. If we call the plane's acceleration $a_p$ then the total acceleration is:

$$ a = g + a_p $$

and the weight becomes:

$$ W = m(g + a_p) $$

At the top of the loop the plane is accelerating downwards so $a_p \lt 0$ and that means $W \lt mg$ i.e. your weight is reduced.

Though I've never done it I'm told it is possible to pull a loop so tight that you experience negative acceleration at the top of the loop i.e. the plane's acceleration downwards is greater than the gravitational acceleration upwards and your weight becomes negative. I would guess this is what the post on the Aviation SE means.

  • $\begingroup$ Ideally a loop is a circle where you pull about 2g at the bottom to start the upward curve, then at the top you could pull almost as little as 0g, progressing back to 2g at the bottom. You could make it tighter, with 3g at the bottom and 1g at the top. Or you could make it looser, with 1.5g at the bottom and -.5g at the top, meaning at the top you are in negative g. Extended negative g requires aerobatic capability. $\endgroup$ – Mike Dunlavey Mar 30 '18 at 13:19

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