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If your angle of attack in landing pattern is 14 degrees but your vertical velocity is 600 feet/minute, does the HUD show you 1G or not? Why?

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I presume, an implicit assumption in your question is that we are flying a perfect glideslope, that is, a straight line with a constant speed, and without disturbances.

If this is the case, the acceleration will be zero by definition, and the only effect we (and an accelerometer) will feel is gravity.

The 'G-meter' shows the acceleration (incl. gravity) with respect to a calibrated standard 1G. Now, the question is, will gravity be the same from the point of view of the device/pilot?

The answer is: almost, but not quite.

First of all, the 'G-meter' typically shows the normal acceleration: that is, along the body Z axis. It is 'vertical' for the pilot, but not necessarily vertical for the Earth. This is the critical acceleration component, the greatest one (because the wing can produce by far the greatest force on an airplane), the one that is always specified in the flight manual and the one that you might exceed in normal manoeuvring. This is why the device shows it and not the 'total' combined acceleration.

Thus, the gravity component as measured by the 'normal' accelerometer will depend on the airplane attitude. In terms of G, the factor will be $cos(pitch) \cdot cos(roll)$. For the given example, if we assume the glideslope angle -3°, pitch will be $14-3 = 11°$, and with zero roll, the measured G will be $cos(11°) \approx 0.98$. Very small difference indeed.

However, in a more extreme example of a vertical dive, the accelerometer will show zero (0G).

But that's not all. If we want to be really pedantic and measure beyond 1%, we'll need to take into account that the apparent gravity changes depending on the location. Given that the accelerometer shows, essentially, weight of a calibrated mass, it will indicate different G even if the aircraft is standing on the ground.

I'm not talking about gravity anomalies, which are truly miniscule. The biggest contributor is Earth rotation, which adds some centripetal acceleration, the more of it the closer to the equator you get. At the equator, it reduces your weight by almost $0.4$% with respect to the pole.

A further $0.1$% can be accounted by the fact that Earth radius is greater at the equator than at the poles - by about 21 km. This is typically a bigger contributor than the aircraft's own altitude (esp. given that glideslopes are usually near the ground).

Altogether, if we calibrated at a pole and then make an approach near the equator in the stated conditions, we can expect almost $2.5$% reduction of the indicated G-force.

In truth, the calibration is usually done for middle latitudes (and digital AHRS may calibrate before takeoff on the actual location), so the difference will be even smaller. But anyway, the biggest contributor - attitude - is unaffected by these small changes. Yet, it is still small by itself, often below the accuracy of the G-meter.

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  • $\begingroup$ " Thus, the gravity component as measured by the 'normal' accelerometer will depend on the airplane attitude. In terms of G, the factor will be cos(pitch)⋅cos(roll). For the given example, if we assume the glideslope angle -3°, pitch will be 14−3=11°, and with zero roll, the measured G will be cos(11°)≈0.98. Very small difference indeed."I believe this pitch will be 14+3=17° real AOA into the wind $\endgroup$ – George Geo Jul 13 at 9:30
  • $\begingroup$ @GeorgeGeo Landing pitch is 'up', glideslope pitch is 'down' so 14-3 is correct. Either way, 17° pitch also only makes for a factor of 0.95 so hardly significant. $\endgroup$ – Sanchises Jul 13 at 13:17
  • $\begingroup$ You said that the above horizontal line is 11 .Now I understand that. But my view is the total 17 degrees AOA because you sink . $\endgroup$ – George Geo Jul 13 at 17:47
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    $\begingroup$ @George, but you specified that AoA is 14°. AoA, by definition, is with respect to air, not ground. Glideslope always goes down (from the aircraft), so we are descending, thus air comes a bit from below. For a given AoA, the pitch attitude will be lower (but still nose up). $\endgroup$ – Zeus Jul 14 at 0:23
  • $\begingroup$ Corect , touching down with the main landing gear first. $\endgroup$ – George Geo Jul 14 at 6:42
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the G meter measures acceleration. If you are descending at constant vertical speed or climbing at constant vertical speed, it will read 1G.

The difference in the force of gravity between that at the earth's surface and that at 35,000 feet is far too small to be detected by an aircraft's G meter.

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    $\begingroup$ This is correct. It’s important to mention that a G meter measures accelerative force relative to the gravitic constant of earth, not actual movement. Sitting on the ground, entirely immobile, you would measure 1G. Accelerating toward the earth at 9.8 meters per second per second (m/s/s) would display 0G, or ‘relatively’ weightless. Accelerating away from earth at 9.8 m/s/s, represents a doubling of the standard accelerative force or 2G even though you are only accelerating upward at the same speed that gravity would cause you to fall. $\endgroup$ – Aaron Holmes Jul 11 at 21:12
  • $\begingroup$ @Aaron, the problem is, the 'G-meter' typically displays only the normal acceleration (or rather, force with respect to gravity, as you rightly say). Thus the indication will depend on attitude. $\endgroup$ – Zeus Jul 13 at 1:14
  • $\begingroup$ Ok. Yes. Maybe it's splitting hairs, but @Zeus is correct. It only measures the intended force perpendicular to the plane in which it is oriented - pragmatically speaking, perpendicular to the seat cushion under your butt. $\endgroup$ – Aaron Holmes Jul 13 at 1:44
  • $\begingroup$ But let's do an 4 point aileron roll in a level flight. Banking to the left 90 degrees the a/c will show you 0.5 G's then upside down - 1 G $\endgroup$ – George Geo Jul 13 at 7:16
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    $\begingroup$ @George, banking 90° (and stopping there for a bit) will show 0G. Inverted [level] flight is indeed about -1G. Climbing 45° (normally), assuming an additional AoA ~8°, will show cos(53°) ~ 0.6 indeed. $\endgroup$ – Zeus Jul 14 at 0:30
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If you fly straight ahead and at constant speed (vertical and horiz) then the only acceleration on the airframe will be the gravity, so total 1G as you normally understand it.

Wether the instrument will actually display 1G depends on calibration (funny trivia, gravity acc. at 35'000 feet is less than 1G since you are further from the Earth) and also depends if instrument shows total acceleration or longitudinal/vertical components.

To put this another way, just because you are descending it will not show <1G, just as it will not show >1G if you are climbing. If you think about it, it would violate a lot of Einstein's relativity postulates as you would be basically be able to tell how you are moving in relation to Earth by using a simple self contained accelerometer.

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