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Finding the mach number of an aircraft

  • Thread starter jmm5872
  • Start date
  • #1
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A supersonic aircraft is flying parallel to the ground. When the aircraft is directly overhead, an observer sees a rocket fired from the aircraft. 10.4 s later the observer hears the sonic boom, followed 3.37 s later by the sound of the rocket engine. What is the mach number of the aircraft?

I think I was too simple in my assumptions, but I used trig to find the half-angle of the mach cone. I used the adjacent side as 10.4, and the opposite side as 13.77, and took the inverse tan. I got an angle of 52.9375.

Then I used this to find the mach number by using this equation:

Mach # = 1/sin([tex]\theta[/tex])

=1/sin(52.9375) = 1.25317
 

Answers and Replies

  • #2
1,925
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Those 2 sides are not the 2 sides next to the 90 degree angle, so taking an inverse tan was wrong.

I calculated the distance from the point from where the sonic boom is first heard to the point where the rocket is released with pythagorars. (the other 2 sides are 13.77c and 10.4c where c is the speed of sound).
This distance is covered by the plane in 3.37s, so it's equal to 3.37v, where v is the speed of the plane.
 
  • #3
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I'm still a little confused by this. I am trying to set the triangle up and can't get past this diagram...I know you said it was wrong, but I can't picture the triangle you are describing.

I tried to attach a drawing I made, I hope it works
 

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  • #4
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Your triangle is actually Ok, (and mine wasn't) but to get [itex] \tan{\phi} [/itex] you need to find the length of the sides in meters, and not in seconds.
The vertical distance is [itex] 13.77 v_{sound} [/itex] and the horizontal distance is [itex] 10.4 v_{plane} [/itex]

When you have an expression for [itex] \tan{\phi} [/itex], you can combine that with

[tex] \frac { v_{plane} } { v_{sound} } = \frac {1} {\sin{\phi}} [/tex]
 

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