Centrifugal force over head? Come on. The Centrifugal force overhead is less than gravity. Guess I never learn simple physics.
'not sure whether Robert was pulling our legs, but... Let's take a look. Of course, a model's speed is reduced in climb: so let's look at both ends.
First, flying 5-sec laps on 70' lines equates to about 88 ft/s or 60 mph. If that speed
were maintained directly overhead in a wingover, the Centripetal acceleration would be...
V
2/R = 88
2/70 = 110.63 ft/sec
2Since the acceleration of gravity is 32.2 ft/sec
2, that's 110.63/32.2 = 3.4 g's.
So the CF (centri
fugal force = - centripetal force) on the lines at the top of the circle would be 3.4 times the weight of the model.
Subtracting weight from line tension overhead (3.4 g's - 1.0 g's = 2.4 g's), a 60-oz model would then have a line tension of 2.4 x 60 oz = 144 oz, even without aerodynamic or thrust effects.
In order for the weight to just equal the CF, the plane's speed will have to have been reduced during climb to 37 mph, or 54% of it's level-lap value.
That's because for V
2/70' to equal 32.2 Ft/sec
2, v = square root of (32.2 x 70) = 47.48 ft/sec.
So a successful stunter will need to be flying faster than this.
Three degrees of out-thrust would only add (sin 3
o =) 5% of the thrust to line tension, if it actually
did produce 3 degrees of out-thrust, after altering the yaw angle. It would reduce CF by about 0.3%.
So CF far overshadows anything out-thrust could provide. If the plane is flying only at marginal speed, what little it adds might help one over the top. If anyone wants to quantify thrust here, be my guest.
SK
Edited to remove dumb error(s).