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To: Darth Reagan
Wouldn't you think turning the plane at 450mph would be about the same amount of force on the tail section if not more? Do we have an engineer out here who knows this answer?
10 posted on 11/16/2001 1:19:07 PM PST by samuel_adams_us
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To: samuel_adams_us
Perhaps .8Gs at a direct, perpendicular angle to the tail is rare, I don't know. I'm not questioning based upon some gut assumption. Let's hear from the engineers. If the NTSB is nuts, some pointy-heads at universities will surely tell us.
21 posted on 11/16/2001 1:19:08 PM PST by Darth Reagan
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To: samuel_adams_us
Wouldn't you think turning the plane at 450mph would be about the same amount of force on the tail section if not more?

Depends on how the turn is made. A properly coordinated turn requires very little continuous rudder force, because what you are actually doing is changing the lift vector through the center of the plane to the left or the right.

Think of a plane dangling on a string, with it attached at the center of gravity so that the wings and fuselage are level. If you tilt that string to the left or right (by banking the wings) while keeping the string perpendicular to the wings/fuselage and pulling on the string to maintain altitude, the plane turns left or right.

Maybe someone can explain it better in words. It's hard for me to do without pictures.

36 posted on 11/16/2001 1:19:09 PM PST by justlurking
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To: samuel_adams_us
As your sitting in your chair reading this, place your hands on your desk and push yourself to a standing position. This my friends is equivlent to 1.42 G's (centralized) of force transfereing from your palms to your desk. Lord help us if our planes can't handle a little push like that
44 posted on 11/16/2001 1:19:10 PM PST by westnews
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To: samuel_adams_us
Wouldn't you think turning the plane at 450mph would be about the same amount of force on the tail section if not more? Do we have an engineer out here who knows this answer?

Turning a plane at 450mph can cause 0 to infinite G's depending on the rate of change in direction.

52 posted on 11/16/2001 1:19:10 PM PST by biblewonk
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To: samuel_adams_us
Consider this: If you could turned the plane on its side so that the tail fin is horizontal to the ground, you would be exerting 1.0G on the fin. In other words, this is hogwash
152 posted on 11/16/2001 1:19:30 PM PST by hound
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To: samuel_adams_us
Wouldn't you think turning the plane at 450mph would be about the same amount of force on the tail section if not more? Do we have an engineer out here who knows this answer?

Maybe if you stomped on the rudder at 450mph, maybe. However you don't turn an aircraft that way. You do turn most missles that way, it's called skid to turn. With an aircraft, and some missles, cruises for example, you bank the beast to turn it. That way you use the lift force from the wings, i.e. positive Gs in the Up direction of the aircraft, rather than side force, which is hard to generate with that little bitty rudder, and hard on that rudder too. When you bank to turn, you use the rudder to keep the turn "coordinated", that is to minimize those side forces, which just cause more drag, which you don't want. You must also pull back on the stick or yoke, to increase the angle of attack so as to produce the extra lift force that you need to drag yourself around the turn, while still retaining enough of a verticle (earth axis not aircraft axis) component to hold the airplane up against gravity. For example a 45 degree banked turn requires the aircraft to pull 1.414 (sqaure root of 2) G's parrallel to your spine, to keep the aircraft flying level, but turning. You'd get one "G" of "force" vertically and one horizontally. That one "G" of "force" would cause you to turn about 2.8 degrees per second at 450 mph. At a lower speed the same accelleration would be produce a faster turn rate, because the turn rate is inverserly proprotional to the speed and directly proportional to the horizontal accelleration.

psi-dot = 180/Pi * A / V

where psi-dot is the turn rate in degrees per second. A is the accelleratin and V is the speed. Speed and accelleration must be in compatible units, such as ft/sec/sec and ft/sec or meters/sec/sec and meters/second. The turn rate comes out in degrees per unit time, if you want to use some other measure of time, minutes, hours, days or fortnights. Leave off the 180/Pi if you want the rate in radians per second.

Never try to do engineering calculations with furlongs as the measure of length and fornights as the measure of time. You'd be sure to blow a gasket somewhere along the way, and your bridge would surely fall or your airplane crash. :).

425 posted on 11/16/2001 1:22:08 PM PST by El Gato
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