Originally posted by mlprior
When you are traveling at an airspeed of about 180 knots (or say 300 km/hr) for our calculation purposes, and you enter a 60 degree angle of turn. How many g's are you in?
Assuming you are holding altitude and airspeed constant through a 360 degree turn.
This is something that I think about often....S
I don't think its quite worded right. The equations for centripital force (the outward force) needs the radius of the turn, which you can probably derive from the given data, at least I think it can, but 'you enter a 60 degree turn' doesn't sound like enough data. So you can make an assumtion or two to get the radius I think. For instance, you make a 60 degree change of direction but how many seconds does that take? And is it a 60 degree turn or a 360? If you know the radius of the turn, you can think about this:
You are at some point in time going to reach the 90 degree point, so that represents stopping in one dimension and starting to move in another dimension, stopping the forward flight and starting a left to right acceleration. Does that make sense? Thats why you need the radius or the time it takes to do that. If you start into a turn and do 90 degrees in one second and you are going 300 Km/hr (83 meters per second) then you are effectively stopping in one second, or 83 meters per second per second which is about 8 G's. If it takes 10 seconds to make the same turn, then you are doing 1/10th of that or 0.8 G.
See why you need more information? (1 G is 9.8 meters per second per second which you can round off to 10 for sure to make the calc easier, so 83/9.8 is 8.5 G's, assuming the change takes place in one second.
If it takes 2 seconds then it's just 8.5/2 or 4.25 G's so you really need that time)