For the eggheads: The math behind an uncoordinated turn
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Listen to bubbers he is right on the money with this stuff.
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We all know a side slip is required any time you use the side slip to land so the actual added speed is dependant on how much side slip you use. Sometimes you use a little, sometimes a lot, so adjust for conditions on short final. Don't get any charts out just land and use your eyeballs. Always worked for me. Flying isn't that hard if you keep it simple.
Simple situations can be made hard by making it complicated as we are here.
Simple situations can be made hard by making it complicated as we are here.
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Well, bubbers44, that's what I sorta expected.
Almost.
In my T-craft, runway length is hardly a problem, but obstacle clearance at local fields can be a challenge. So I slip with the nose well down to kill some L/D - I couldn't tell you my IAS if I had to, my eyeballs are outside the plane. Gotta do it that way, got no RA, and I un-slip it at about 10 feet.
Many years ago my dad test-flew a rebuilt J-3 for a friend. After a half-hour airwork, he came back and landed first time using no more than 500 feet of runway. Although he had plenty of prior J-3 time, that had been 20 years earlier.
But thanks for the "big iron" perspective.
Almost.
In my T-craft, runway length is hardly a problem, but obstacle clearance at local fields can be a challenge. So I slip with the nose well down to kill some L/D - I couldn't tell you my IAS if I had to, my eyeballs are outside the plane. Gotta do it that way, got no RA, and I un-slip it at about 10 feet.
Many years ago my dad test-flew a rebuilt J-3 for a friend. After a half-hour airwork, he came back and landed first time using no more than 500 feet of runway. Although he had plenty of prior J-3 time, that had been 20 years earlier.
But thanks for the "big iron" perspective.
Last edited by barit1; 7th Dec 2012 at 11:58.
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Hi barit1,
He could "push the beast's" rudder until he was about to run out of aileron control, which is easy to feel in a Boeing through the control wheel displacement.
If he were trying to do the same thing in an Airbus FBW - then he'd have to be looking at the Flight Control System Page to see how much aileron was being applied for him.
Was he merely lucky not to depart the regime of positive roll control, or did he have some insight beyond his seat-of-the-pants about how far to push the beast?
If he were trying to do the same thing in an Airbus FBW - then he'd have to be looking at the Flight Control System Page to see how much aileron was being applied for him.
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I flew the T craft and taught my FAA medical examiner how to fly in it and when I got really bored one day found I had flown over 75 types of aircraft going through my log book including experimental biplanes which I certified for aerobatics for the FAA. I was lucky when aviation was not so restrictive we could do this. My FAA friend had a one of a kind mini biplane that was a blast but the only one built on some pilots garage with drawings on the floor. It was called the Bailey Bitty Bipe.
Doing slips in any airplane is easy, just push rudder one way and aileron the other. You can call it a forward slip or a side slip depending what you are using it for. We learned to do it before solo 50 years ago.
It still works today. I feel sorry for the new guys who are taught automation and didn't get to experience what I did.
Doing slips in any airplane is easy, just push rudder one way and aileron the other. You can call it a forward slip or a side slip depending what you are using it for. We learned to do it before solo 50 years ago.
It still works today. I feel sorry for the new guys who are taught automation and didn't get to experience what I did.
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With reference to Superpilots diagrams..
For a correct turn with no loss of height..
Vertically..
mg = LCos(x)
Where
m=mass of aircraft
g = gravity
L = Lift
x = bank angle
Solve for L
L = mg/Cos(x) ...................... (1)
Horizontally..
mv^2/r = LSin(x)
where
v = velocity
r = radius of turn
Solve for L
L = mv^2/rSin(x) ......................(2)
Equate (1) and (2)
mg/Cos(x) = mv^2/rSin(x)
Mass cancels
rearrange
Sin(x)/Cos(x) = v^2/rg
Sin(x)/Cos(x) = Tan(x)
so I think in a normal turn
Tan(x) = v^2/rg
If Tan(x) > v^2/rg then it's a slip
If Tan(x) < v^2/rg then it's a skid
Someone check the maths :-)
For a correct turn with no loss of height..
Vertically..
mg = LCos(x)
Where
m=mass of aircraft
g = gravity
L = Lift
x = bank angle
Solve for L
L = mg/Cos(x) ...................... (1)
Horizontally..
mv^2/r = LSin(x)
where
v = velocity
r = radius of turn
Solve for L
L = mv^2/rSin(x) ......................(2)
Equate (1) and (2)
mg/Cos(x) = mv^2/rSin(x)
Mass cancels
rearrange
Sin(x)/Cos(x) = v^2/rg
Sin(x)/Cos(x) = Tan(x)
so I think in a normal turn
Tan(x) = v^2/rg
If Tan(x) > v^2/rg then it's a slip
If Tan(x) < v^2/rg then it's a skid
Someone check the maths :-)
Last edited by cwatters; 8th Dec 2012 at 10:32.
