# How far could the Wright brother's catapult have thrown their unpowered plane?

I know that the Wright brothers started to use a catapult in September 1904 to help their Flyer II take off easily even when no headwind was available.

Assuming the propellers did not turn, or they were not even mounted on their shafts, how far away the catapult could have thrown Flyer II together with its engine and pilot? 1m, 5m, 10m, 25m, 100m or more? Even a rough evaluation, based for example on an experiment with a scale model, would be good for me.

(Note: I am not interested in a pure ballistic evaluation, considering the plane the equivalent of canon ball of mass, m, that is accelerated to the speed, v, and launched at an angle alpha.)

• I'm not sure that this experiment has been done with a scale model, are you actually asking that somebody build an accurate scale model and attempt to replicate this for your answer? Otherwise a purely ballistic answer or mathematical one is all you can hope for... – Ron Beyer Mar 13 '17 at 2:41
• The catapult used 1400lbs of weights falling from a 20-foot tower for propulsion. Maybe somebody can do the math and figure out how much energy that imparts vs the performance of the Flyer – TomMcW Mar 13 '17 at 3:04
• It appears that a Flyer II, with the engine stopped and accelerated only by that falling weight, could not have flown at all because the energy communicated by the weight to the plane is about the same as that required by the 925 lbs Flyer II to reach the take off speed (30 mph). --- E_weight=(1400 pound) * (20 ft) * (9.81 (m / (s^2))) = 37 975.87 joules --- E_plane_take_off=(925 pound * ((30 mph)^2)) / 2 = 37 732.25 joules. The moment the take off speed is attained, the weight stops, the aerodynamic friction diminishes the speed and the plane remains below the required 30 mph flight speed. – Robert Werner Mar 13 '17 at 5:45
• "thrown" is not the right word - it merely pulled the plane down a track for a distance. If the plane didn't get enough airspeed from the pull and the headwind, it wouldn't take off when the elevator was manipulated. – CrossRoads Jun 6 '18 at 19:15

$$Mw*Hw*g=1/2Mf*Vf^2+1/2Mw*Vw^2$$ $$635kg*5m*9.8m/sec^2=1/2*408kg*Vf^2+1/2*635kg(Vf/3)^2$$ $$130m^2/sec^2=Vf^2$$ $$Vf=11.4m/sec=25mph$$ Since the Flyer II needed 28-30mph to become airborne, the catapult cannot launch the unpowered flyer by itself, even excluding friction and drag.