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Varsity
Flown Subs, Ground Bounce, and People in 1/2 space
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<blockquote data-quote="Charles Harrigan" data-source="post: 24887" data-attributes="member: 23"><p>Re: Flown Subs, Ground Bounce, and People in 1/2 space</p><p></p><p></p><p> </p><p>Imagine two right triangles: one formed by the normal from the ground to the sub array, the ground and the particular vector we are interested in. The second by you, the ground and the reflected vector</p><p></p><p>[sub array]</p><p> | \ </p><p> | \ </p><p> | \ </p><p> | \ </p><p> | \ </p><p> | \ </p><p> | \ </p><p> | \ </p><p> | \ ☺ </p><p> | \ /‒|‒ </p><p>|____________ \/ _ᴧ_______[ASCii by chuck]</p><p></p><p></p><p>The ratio of the distance between your ear height divided by the distance to the point of incidence is equal to the ratio of the height of the sub array divided by the distance from the intersection of the normal to the ground plane to the point of incidence. Then once you have those distances figured out, you can use the Pythagorean theorem to determine the distance traveled by the reflected wave(find the hypotenuse of the two triangles and add them). You'd use only need pythagoras to determine the distance the direct wave traveled. This is of course assuming there is no refraction at the air ground boundary. You could get tangent involved and write an equation that relates the difference in distances. I'm estimating that if you did write that equation and relate that to the amount of summation and cancellation, for one particular frequency, the level over the distance would vary sinusoidally. (and some people say physics, diffeq and vector calculus is useless....)</p></blockquote><p></p>
[QUOTE="Charles Harrigan, post: 24887, member: 23"] Re: Flown Subs, Ground Bounce, and People in 1/2 space Imagine two right triangles: one formed by the normal from the ground to the sub array, the ground and the particular vector we are interested in. The second by you, the ground and the reflected vector [sub array] | \ | \ | \ | \ | \ | \ | \ | \ | \ ☺ | \ /‒|‒ |____________ \/ _ᴧ_______[ASCii by chuck] The ratio of the distance between your ear height divided by the distance to the point of incidence is equal to the ratio of the height of the sub array divided by the distance from the intersection of the normal to the ground plane to the point of incidence. Then once you have those distances figured out, you can use the Pythagorean theorem to determine the distance traveled by the reflected wave(find the hypotenuse of the two triangles and add them). You'd use only need pythagoras to determine the distance the direct wave traveled. This is of course assuming there is no refraction at the air ground boundary. You could get tangent involved and write an equation that relates the difference in distances. I'm estimating that if you did write that equation and relate that to the amount of summation and cancellation, for one particular frequency, the level over the distance would vary sinusoidally. (and some people say physics, diffeq and vector calculus is useless....) [/QUOTE]
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