G and its real meaning
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G and its real meaning
Anyone care to help me out with the relationship between accelerations (lateral and vertical)and other parameters with which your average joe may be more familiar?
For example what sort of v/s would equate to a 2.0g touchdown?Or put a high speed RTO in g terms.I appreciate that the relationship is not a linear one but anything on this type of stuff would be welcome.Checkboard - you may be the man.
Thanks in advance y'all.
For example what sort of v/s would equate to a 2.0g touchdown?Or put a high speed RTO in g terms.I appreciate that the relationship is not a linear one but anything on this type of stuff would be welcome.Checkboard - you may be the man.
Thanks in advance y'all.
Guest
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Acceleration is the rate of change of velocity.
Gravity gives an acceleration of 32 ft/sec/sec
An aircraft doing 150kt (no wind) on a 3 degree glideslope has a vertical velocity of 13.3 ft/second.
The formula we need is
v squared = u squared + 2as
where u is initial velocity
v is final velocity
a is acceleration
s is distance travelled between u and v
We have u=13.3, v=0
All the vertical deceleration occurs in compression of the oleos and tyres. Say a movement of 3 feet. Assume the load/deflection curve is uniform over that 3 feet, and oleos do not hit limits.
Then a = u squared / 2s = 13.3* 13.3 / (2 * 3) = 29.5 = 29.5/32g = .92 extra gs
Note the effect of the v squared
If you can at the last moment halve the vertical speed to say 6.6 ft/sec while the oleos are deflecting, you get only 0.23 extra g.
Gravity gives an acceleration of 32 ft/sec/sec
An aircraft doing 150kt (no wind) on a 3 degree glideslope has a vertical velocity of 13.3 ft/second.
The formula we need is
v squared = u squared + 2as
where u is initial velocity
v is final velocity
a is acceleration
s is distance travelled between u and v
We have u=13.3, v=0
All the vertical deceleration occurs in compression of the oleos and tyres. Say a movement of 3 feet. Assume the load/deflection curve is uniform over that 3 feet, and oleos do not hit limits.
Then a = u squared / 2s = 13.3* 13.3 / (2 * 3) = 29.5 = 29.5/32g = .92 extra gs
Note the effect of the v squared
If you can at the last moment halve the vertical speed to say 6.6 ft/sec while the oleos are deflecting, you get only 0.23 extra g.
