Is it possible to calculate a flight path of a plane using dB and duration it was heard? Is it possible to calculate the flight path past a boat of an aeroplane based on the following information?
The plane was a prop plane which can be heard at 88 dB at 1000 ft.
The plane was cruising at 149.6 mph at 10,000 ft.
The plane was heard for 1 minute.
The plane was heard off the starboard quarter off the boat.
Thanks
 A: The only thing you can calculate from this is the distance of closest approach of the plane from your location (in other words, how far to the starboard it was), and even then, without knowing how loud the background noise was (e.g. Did you have music on? Did you have the engine on? Were you on the ocean, and near enough to the shore to hear waves breaking? etc.), this number will be very, very imprecise. We also have to assume that the plane is traveling in a straight line at the same altitude for the entire minute. 
As a rule of thumb, a doubling of the source distance results in a loss of 6 dB of sound level. Assuming an ambient background noise of around 60 dB (source: http://www.industrialnoisecontrol.com/comparative-noise-examples.htm and assuming there's some background music or conversation), the plane's engine noise would disappear into the background after 4.7 doublings (going from 88 dB to 58 dB), at a distance of 25,400 feet. Given that we know that 10,000 feet of that distance is vertical height, you can only hear the plane when its ground position is within a circle of radius $\sqrt{25,400^2-10,000^2}=23300$ feet from you.
The plane is traveling at 149.6 mph (219.4 ft/s) for one minute, so it travels 13160 ft while you're able to hear it. This means the plane is traversing a chord of the circle with length 13160 ft, as shown in this diagram:

Fortunately, there's actually a fairly simple formula to calculate the distance of closest approach $d$ from the length of the chord $a$ and the radius of the circle $R$ (from http://mathworld.wolfram.com/CircularSegment.html):
$$d=\frac{1}{2}\sqrt{4R^2-a^2}$$
Plugging in $a=13160$ ft and $R=23300$ ft, we get $d=22400$ ft. So, at its closest, the plane was 10,000 ft up and roughly 22,000 ft to the side somewhere.
Note that this number depends heavily on the things that weren't precisely measured, especially the background noise. For example: 


*

*if the ambient noise were 65 dB instead of 60 dB, then the radius of the circle would be only 10,000 ft, with the distance of closest approach changing to 7500 ft.

*if the ambient noise were 66 dB instead of 60 dB, then the radius of the circle would be only 7800 ft, with the distance of closest approach changing to 4240 ft.

*if the ambient noise were 55 dB instead of 60 dB, then the radius of the circle would be 44,135 ft, with the distance of closest approach changing to 43600 ft.


So if we don't know how loud the ambient noise is to within 5 or 6 dB, then we don't know whether the plane was 4,000 ft to the side or 40,000 feet to the side.
A: 
The plane was a prop plane which can be heard at 88 dB at 1000 ft. The
  plane was cruising at 149.6 mph at 10,000 ft. The plane was heard for
  1 minute. The plane was heard off the starboard quarter off the boat.

This is likely leading to a point about the Transit system. The key to solving this is that they have provided a speed, a time, and a vague location.
Think about the plane as it passes by. The pitch of the aircraft engine will change due to the doppler shift. If it passes directly overhead, it will be shifted up in frequency as it approaches, and down as it flies away.
In most illustrations of doppler shift they simplify things by assuming the object is travelling right at you. This is not the case for an aircraft that passes overhead, because it never gets closer than it's altitude. In this case the actual rate it approaches you depends on its location. For instance, if the aircraft is flying at 500 and is 10 miles away and coming at you, it's actual line-of-sight speed is close to its actual speed. But when it's right overhead, it's not approaching you at all, so its line-of-sight velocity is zero, and it's doppler shift drops to zero.
So if you think about the entire 1 minute pass, you'd hear something like "beeee-aaaaa-ohhhh-uuuuu".
Now here's the thing... if the aircraft does not pass directly overhead, but off to one side for instance, then the change in the line-of-sight velocity will always be lower. Do you see why? So that sound you would hear if it passed overhead will sound different if it's a mile away to the side. The overall pattern of the sound is the same, but the exact speed that the frequencies change will be different. That that exact pattern directly reveals the distance to the object.
Now that distance can be on either side of you, which is why they add the "starboard quarter" bit.
This is how the Transit system worked, except the sound was a radio transmitter, and the orbital path was known to a high degree. To figure out which side of you it was on, you had to feed in an initial point, typically using some other system like an inertial navigation unit or even just dead reckoning. Now you listened as the satellite passed, calculated the exact distance of its closest approach, calculated where the satellite was at that time, and presto.
Navstar does not work this way, it uses clock comparison.
