09 Jul 08
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....
09 Jul 08
Originally posted by mlpriorI counted five but I might have missed one or two.
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....
09 Jul 08
2 edits
Originally posted by mlpriorI 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:
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
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)
09 Jul 08
Originally posted by sonhouseI can always count on you, Sonhouse to make things so much more complicated than they actually are.
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 deg ...[text shortened]... d.
If it takes 2 seconds then it's just 8.5/2 or 4.25 G's so you really need that time)
😉
I could specify a little more in this way. The angle of the wing to the horizon is 60 degrees. You are doing a turn around a point in space starting at north and ending up again at north (360 degrees). Your airspeed indicates 300 km/hr.
Does that help?
09 Jul 08
Originally posted by mlpriorIt doesn't matter, you can always analyze it in terms of how many seconds does it take to turn 90 degrees, thats the key. That's not exactly rocket science, just freshman physics.
I can always count on you, Sonhouse to make things so much more complicated than they actually are.
😉
I could specify a little more in this way. The angle of the wing to the horizon is 60 degrees. You are doing a turn around a point in space starting at north and ending up again at north (360 degrees). Your airspeed indicates 300 km/hr.
Does that help?
09 Jul 08
Originally posted by mlpriorWithout calculation, I think 2 g would be reasonably.
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....
With calculation: 1/cos(60) is exactly 2. I wonder if this is right.
09 Jul 08
Originally posted by FabianFnasLike I said, if it takes one second to change angles by 90 degrees at 300Km/hr, it's 8.5 G's. If ten seconds, 0.85 G's. Not exactly Phd level stuff here.
Without calculation, I think 2 g would be reasonably.
With calculation: 1/cos(60) is exactly 2. I wonder if this is right.
10 Jul 08
Originally posted by mlpriorThe angle of your wings to the horizon doesn't really tell you how fast you are turning; you could be sliding laterally on the air as well.
I can always count on you, Sonhouse to make things so much more complicated than they actually are.
😉
I could specify a little more in this way. The angle of the wing to the horizon is 60 degrees. You are doing a turn around a point in space starting at north and ending up again at north (360 degrees). Your airspeed indicates 300 km/hr.
Does that help?
You really do need how many degrees/second or radius of the turn to tell you what centripetal acceleration you are experiencing.
11 Jul 08
Originally posted by FabianFnasYou are absolutely correct Fabian!
Without calculation, I think 2 g would be reasonably.
With calculation: 1/cos(60) is exactly 2. I wonder if this is right.
A 60 degree turn will pull a load factor of 2 g.
Load factor = lifft/weight = 1/cos(bank angle) = 1/cos(60)
Exactly as you stated!
😉
11 Jul 08
Originally posted by mlpriorYezz!
You are absolutely correct Fabian!
A 60 degree turn will pull a load factor of 2 g.
Load factor = lifft/weight = 1/cos(bank angle) = 1/cos(60)
Exactly as you stated!
😉
But I don't understand where "an airspeed of about 180 knots" or "through a 360 degree turn has to do with anything...? That confused me and therefore I didn't know if my answer was right in this situation...
But 1/cos(angle) does work in various situations:
Bicycle or mtorcyckle turn, where you 'bank' in relation to your turn.
Various attractions at the amusement parque.
Even when you run through narrow corridors at the office. Well, 60 degrees 'banking' perhaps is not very usual, but the same formula applies.
What about USS Enterprice in StarTrek banking 60 degrees in open space to make an areodynamically correct turn...?
11 Jul 08
Originally posted by FabianFnasThis does not seem right to me. If you are driving around a bank of 60 degrees, you say the load is 2 G's but it still seems velocity dependent. If you are going 2 Km/hr you would get a lot less G load than if you are going 200 Km/hr. If the 2 G load is correct it looks like it takes 17 seconds to complete 360 degrees.
Yezz!
But I don't understand where "an airspeed of about 180 knots" or "through a 360 degree turn has to do with anything...? That confused me and therefore I didn't know if my answer was right in this situation...
But 1/cos(angle) does work in various situations:
Bicycle or mtorcyckle turn, where you 'bank' in relation to your turn.
Various attra ...[text shortened]... n StarTrek banking 60 degrees in open space to make an areodynamically correct turn...?
11 Jul 08
Originally posted by sonhouseNot if you're right on the wings. If so then the velocity doesn't matter.
This does not seem right to me. If you are driving around a bank of 60 degrees, you say the load is 2 G's but it still seems velocity dependent. If you are going 2 Km/hr you would get a lot less G load than if you are going 200 Km/hr. If the 2 G load is correct it looks like it takes 17 seconds to complete 360 degrees.
The same rule apply to a supersonic attacker, a Concorde in Mach 1.5, a lightweight 2 hp engined hobby flyer, a deltawinged airglider, or even a bumblebee.
11 Jul 08
Originally posted by FabianFnasHow is it different driving around a banked turn at 60 degrees and flying through a banked turn at 60 degrees?
Not if you're right on the wings. If so then the velocity doesn't matter.
The same rule apply to a supersonic attacker, a Concorde in Mach 1.5, a lightweight 2 hp engined hobby flyer, a deltawinged airglider, or even a bumblebee.
11 Jul 08
Originally posted by sonhouseThe motocycle analogy makes a lot more sense to me as to why you could get the centripetal force based on the angle. If you maintain a 60 degree angle to the ground, it will take a determinate lateral force based on gravity to maintain that angle.
How is it different driving around a banked turn at 60 degrees and flying through a banked turn at 60 degrees?
If you're going 2 km/hr while leaning over to 60 degrees, you will turn MUCH faster than if you are going 200 km/hr, but the centripetal force will be the same.
Translating that to an aircraft, I was having difficulty determining the physics needed to maintain a perfect radius at a precise altitude. From experience, I know that angling the wings to the ground does not necessarily cause the plane to turn at all -- instead you will keep the same heading and start sliding sideways (and downward) through the air. I can see that if you were also to keep the same groundspeed (not airspeed) and the same altitude, your angle of turn would determine the centripetal acceleration.
In conclusion, realistically, it would not be feasibile to "try this at home" and get precise results.
