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#1
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not too difficult of a problem but cant seem to figure it out.
the rate of climb is given as 674 feet per minute and the optimal angle of climb airspeed of 78 knots those are the only numbers given, basically vertical 674 fpm horizontal im assuming 78 knots, my friend seems to think it should be tan(a.o.c)=674/78, i keep comming up with 83.4 as an answer and this simply does not make sense, help set me straight or show me where my math is wrong, thanks. |
#2
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You should rewrite the question. It's a mess. In fact, I don't even see a question there.
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#3
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units must be consistent.
674 fpm vertical 78 knots=131.6fps*60s/min=7896ft/m aoc=arctan(674/7896)=4.87 degrees |
#4
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thank you i thought that may be the problem.
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#5
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I would think the airspeed would measure the speed along the hypotenus of the right triangle. The horizontal would be the ground speed. So you would want the
arcsin(674/7896) assuming the knots were converted correctly. Which would give a steeper angle. PairTheBoard |
#6
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technically,it is.
However,the speed is significantly greater than the vertical velocity component that the speed is approximately the horizontal velocity component. you can see that velocity_x=(7896^2-674^2)^.5=7867 ft/min |
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