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G loading v gravity


oldcrusty

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g-bank-angle.jpg

 

g-effect.jpg

 

http://www.aerospaceweb.org/question/performance/q0146.shtml#:~:text=Any%20aircraft%20in%20a%20level,lift%20that%20turns%20the%20aircraft.

 

Check out this website, its all math. As you approach 90 degrees of bank no amount of G will keep your flight path marker level. All of your lift you are generating is going straight out the top of the aircraft and the VV will fall. As to the rate, that looks about right, again, you can do the math but it should be falling at the equivalent of 1 G acceleration straight down if your roll angle is exactly 90 degrees, since that's purely the acceleration of gravity pulling the aircraft down.

 

This math and chart is independent of aircraft type or performance, only limited by how fast the aircraft can go/how many G's the aircraft can actually pull. All works the same for a Cessna, or a hornet, or an F-16, or whatever.


Edited by KlarSnow
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Yea, I'm familiar with the basic concepts. For some weird reason it just occurred to me that the VV might be falling at a faster rate then it should. Heck, after years of flight simming, it's about time I started paying attention to stuff like this and ask questions.:)

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The VV flight path angle is going to evolve according to the trigonometry of the arctangent of the forward speed and the downward speed. Starting at say 500 fps forward and 0 fps downward the VV is on the horizon.

 

Evolving the situation with the assumption of no upward lift at 90 AOB we assume the velocities after 1, 2, 3, 4, etc. sec of 32, 64, 96, 128, etc. fps due to gravitational acceleration. Those correspond to 3.7, 7.3, 10.9, 14.5 degrees.

 

From the video it looks like you're getting almost exactly that ~3.5 degrees per second VV fall. You're doing a little more than 300 knots (500fps) so one expects correspondingly reduced angles.

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