Helicopter ejection seat?

Helicopter ejection seat. Sounds like a terrible idea right? Always mentioned in the same sentence as screen doors in submarines, named as a rhetorical remark to say that something just said is in fact not the worst idea ever. The very thought of the seat brings to mind a pilot being chopped to bits as they head through the whirling helicopter blades, but is that actually what would happen? Or is there a chance that with computerized timing, a person could be pushed through between blade rotations? I will attempt to answer the question with maths.

The first numbers we need are the height of the person as they are pushed through the blades, and the rotational speed of the copter blades at the place the person is ejecting. The first of these is easy to estimate, as a sitting person is approximately 2/3rds of their standing height, that means for a 6 foot person, they would be 4 feet tall when sitting. This will be the number used, except it will be converted to 1.22 meters.

The second number is more difficult to attain. From what I read on the intrablogs, the rotational speed of helicopter blades is around 500 rpm, double because there are two blades spinning at this speed, and you come up with 1000 rpm, which equals 1000/60 rotations per seconds, which is equal to 16.66 rps.

This is where a lot of haphazard guesswork comes into play. Let us use the Bell 206 for our example copter for these equations. This helicopter has a rotor diameter of 10.16 meters with the pilot’s seat roughly 3 feet from the center of the rotor. For arguments sake, let us assume that the width of the pilot requires 50 percent the length of the chord made by the line crossing the circle at three feet from the center of the circle. It takes 06 seconds for each rotation, so the time the pilot has to make it through the blade area is (.06)*(2) seconds=.12 seconds. And because he must travel 1.22 meters in this time (his own height while seated) he must travel at 1.22/.12 m/s=10.1 m/s when he is traveling through the blades. Another estimation puts the blade at 6 feet above the pilots head, assuming the pilot must be traveling at an average of 10.1 m/s while going through the blades, this means he has 6+2=8 ft, or 2.43 meters to accelerate to this speed, by this, and seeing he would only have 5.05/2.43 seconds=2.07 seconds to travel all the way to top speed. This makes acceleration would have to be only 10.1*2.07 m/s^2 so 20.9 m/s^2, plus the 9.8 m/s^2 or normal earth acceleration makes for 3.13g’s.

I can only assume that my calculations are fraught with errors and inaccuracies, but I don’t care all that much. For I have reached a final number through steps that appear logical to my own self.