October 30, 201411 yr More or less. There's a way to use the mach number (2, in this case) to determine the angle of that cone that constitutes the sonic wave, which you need in order to solve the problem. I can't remember how to do it. The sine of that angle is the speed of sound over the speed of the plane, which is the inverse of the mach number. I can't remember how it works. Sounds like a Calculus problem then a simple Trig calculation. LOL It is over my head too, I will sit back and wait for the real experts to take over.
October 30, 201411 yr You gave us the speed of the plane, 1,522.4 mph. That would be A once we do the proper conversion. We need the actual altitude, that would be B. Then C is just plugging and chugging assuming we have the right formula. The problem is that we can't convert it. Distance is velocity multiplied by time and time is what we're solving for. I think I have it. I'm fairly sure that the angle of the sonic wave is 30 degrees. So, we flip the triangle so that the hypotenuse is the distance from the observer to the craft and the height of the plane is the opposite side from the angle. That angle along the hypotenuse remains the same. The tangent of that angle is equal to the ratio of the opposite side divided by the adjacent side. We now have this equation: tan(30) = 300 ft / (2232.85 ft/s x time) The original problem said "a few hundred feet"; I'm going to assume 300 feet. Divide both sides by the tangent of thirty and multiply both side by time (t) and we have... t = 300/(2232.85xtan(30)) t = ~0.23 seconds I'm going to scan my scratch paper and post it. ...wow... I'm really doing just about anything to avoid actually working today.
October 30, 201411 yr The problem is that we can't convert it. Distance is velocity multiplied by time and time is what we're solving for. I think I have it. I'm fairly sure that the angle of the sonic wave is 30 degrees. So, we flip the triangle so that the hypotenuse is the distance from the observer to the craft and the height of the plane is the opposite side from the angle. That angle along the hypotenuse remains the same. The tangent of that angle is equal to the ratio of the opposite side divided by the adjacent side. We now have this equation: tan(30) = 300 ft / (2232.85 ft/s x time) The original problem said "a few hundred feet"; I'm going to assume 300 feet. Divide both sides by the tangent of thirty and multiply both side by time (t) and we have... t = 300/(2232.85xtan(30)) t = ~0.23 seconds I'm going to scan my scratch paper and post it. ...wow... I'm really doing just about anything to avoid actually working today. #TeamRandy
October 30, 201411 yr Cool video. BUT, in all those cases, the plane was ACCELERATING to the point of going the speed of sound and just reaching it in front of the crowds. In mine, the plane has been traveling TWICE the speed of sound for an hour before it gets to you, at a constant velocity. Does that not change the dynamics? I was always under the assumption that being as close to a plane causing a sonic boom would require hearing protection but I didn't notice anyone wearing any.
October 30, 201411 yr The problem is that we can't convert it. Distance is velocity multiplied by time and time is what we're solving for. I think I have it. I'm fairly sure that the angle of the sonic wave is 30 degrees. So, we flip the triangle so that the hypotenuse is the distance from the observer to the craft and the height of the plane is the opposite side from the angle. That angle along the hypotenuse remains the same. The tangent of that angle is equal to the ratio of the opposite side divided by the adjacent side. We now have this equation: tan(30) = 300 ft / (2232.85 ft/s x time) The original problem said "a few hundred feet"; I'm going to assume 300 feet. Divide both sides by the tangent of thirty and multiply both side by time (t) and we have... t = 300/(2232.85xtan(30)) t = ~0.23 seconds I'm going to scan my scratch paper and post it. ...wow... I'm really doing just about anything to avoid actually working today. Somebody had the Esterle brothers for Math and Bob Hublar for Physics at Trinity. LOL
October 30, 201411 yr Somebody had the Esterle brothers for Math and Bob Hublar for Physics at Trinity. LOL And how. Took math from Dennis Esterle three times.
October 30, 201411 yr And how. Took math from Dennis Esterle three times. That should be classified as torture per the Geneva Convention.
October 30, 201411 yr That should be classified as torture per the Geneva Convention. Loved it. At least once a week I think about what I'd have done if I'd have stuck with it. I hated engineering school and wanted nothing to do with building things or learning how to do computer programs or circuits... but I did like the math. There's something really satisfying about putting pencil to paper and just figuring something out.
October 30, 201411 yr Author I was always under the assumption that being as close to a plane causing a sonic boom would require hearing protection but I didn't notice anyone wearing any. Wow...just looked it up. At the source, a sonic boom generates approximately 200 decibels. 140 db's is what's usually specified as being "dangerous" at any sustained level (assuming do hearing protection, bare ears), which is equivalent to being about 100 feet or so from a jet engine. 200 decibels, even for just fraction of a second, if near the source as it was in part of that video...without protection I would say it very well may have caused some hearing damage. Table chart sound pressure levels SPL level test normal voice sound levels pressure sound intensity ratio decibel comparison chart conversion of sound pressure to sound intensity noise sound units decibel level comparison of common sounds calculation
October 30, 201411 yr Bill Engvall talks about flying with the. thunderbirds at low altitude. In his words "I saw bears pooping-- they didn't WANT to...."
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