Nebelwerfer

[karch]

Here is a quick image I made to try and explain my point.

EDIT- changed because I missed the reflection of the line when it crosses the line between the listening posts.

The firing gun is exactly between listening post B and C, so the line drawn would be right down the middle between B and C because ANYWHERE along the line's length, the gun report would reach B and C simultaneously.

The firing gun is, say, 20 units further away from A as it is from B. That doesn't give a distance, but a difference in range of the gun between A and B. The line drawn between A and B is a line that stays 20 units further away from A than from B. Anywhere along the line a sound will arrive at B the same amount of time before it reaches A.

Where the 2 lines intersect is where the gun must be.

Anyone see flaws in my idea or is my description confusing to anyone?

[ February 15, 2002, 06:13 PM: Message edited by: karch ]

[MajorBooBoo]

Originally posted by karch:

Here is a quick image I made to try and explain my point.

The firing gun is exactly between listening post B and C, so the line drawn would be right down the middle between B and C because ANYWHERE along the line's length, the gun report would reach B and C simultaneously.

Anyone see flaws in my idea or is my description confusing to anyone?

Thank you for the drawing.

The case of the simultaneous sound arrival at two receptors has already been discussed. I did this in my Q&A if you go back some pages. It is a special case only. Theres also the case where the sound is not simultaneous but the source and the two receptors are all in a line. In either of these special cases, only direction can be found (ie the gun is somewhere along the line).

You are on on a good track but time is running out. I will type up the answer and post it here today in an hour or two.

Someone also said:

You can not do it with 2 microphones. You would be able to get a possible line, but not a specific point. Think about this logically.

You don't know how far away the gun was fired. If the time differential was 1 one second, that tells you that the distance of the firing gun is 1 second further away from the gun. The speed of sound at sea level is 1115.7 (1100 for math purposes) feet per second.

If the gun was one foot away from microphone A and 1100 feet from microphone B you would get a 1 second delay.

If the gun were 1100 feet from microphone A and 2200 feet from microphone B you would get a 1 second delay.

If the gun were 11,000 feet from microphone A and 12100 from microphone B (a 1,100 ft difference) you would have exactly a 1 second delay/differential..

This person is throwing out the important piece of information about the seperation of the surveyed receptors. This constrainment is vital. Without using it you fall into teh pitfalls that others have been doing here.

use this:

If you have two mikes (x1,y1) and (x2,y2) and a time=distance difference dr12 then you have three equations

r1^2 = (x-x1)^2 + (y-y1)^2

r2^2 = (x-x2)^2 + (y-y2)^2

r2 = r1 + dr12

Use this:

The knowns are:

Speed of Sound

1000 meters between recievers.

404.74 time lag from sound (dr2)

R1=R2-404.74 (or however you make your R's)

exact cartesian coords of recievers

Request: If you could post a image of two concentric circles with points at the center of the two cicles, a point on both the drawn perimeters, I would be very grateful. Make the points all look like a right triangle for the sake of clarity.

The real starting point is to make a correct diagram.

[Thomm]

Originally posted by karch:

I can't show you the math, but by using simple logic and a graph, you can see that 2 microphones will get you a line, and 3 microphones can get you 2 intersecting lines that pinpoint the gun.

Unfortunatly my simple logic is often simply wrong. Anyone agree or dissagree with the above hypothesis?

No, no Scott, you got everything right, just as the others (asok, JonS, ...) before you (including, of course, the historians). In fact, the graphical solution gives the necessary hints to simplify the symbolic calculation, it just has to be translated to the respective algebraic expressions. I do not think this is necessary, though, because there are enough excellent diagrams and statements floating around already!

Regards,

Thomm

P.S.: Cool Homepage!!

[Thomm]

Originally posted by MajorBooBoo:

The real starting point is to make a correct diagram.

The starting AND the ending point for you would be to read and to try to *understand* what all the helpful people wrote in response to your questions! The problem was solved simultaneously by 1) historical research 2) sybolic computation 3) graphical methods. I do not know why you continue to beat the poor dead horse. Must be a psychological experiment ...

Regards,

Thomm

[JonS]

Originally posted by redwolf:

</font><blockquote>quote:</font><hr />

Two point sensors tells you the gun is located somewhere along one of two lines (one in enemy territory, one in your own)

I think it is rather one line leading from your to enemy territory.

</font>

[karch]

Originally posted by MajorBooBoo:

This person is throwing out the important piece of information about the seperation of the surveyed receptors. This constrainment is vital. Without using it you fall into teh pitfalls that others have been doing here.

use this:

If you have two mikes (x1,y1) and (x2,y2) and a time=distance difference dr12 then you have three equations

r1^2 = (x-x1)^2 + (y-y1)^2

r2^2 = (x-x2)^2 + (y-y2)^2

r2 = r1 + dr12

Use this:

The knowns are:

Speed of Sound

1000 meters between recievers.

404.74 time lag from sound (dr2)

R1=R2-404.74 (or however you make your R's)

exact cartesian coords of recievers

Request: If you could post a image of two concentric circles with points at the center of the two cicles, a point on both the drawn perimeters, I would be very grateful. Make the points all look like a right triangle for the sake of clarity.

The real starting point is to make a correct diagram.

I'll try and draw it later tonight, but I think you are mistaken. The ONLY way you can find the radius of your circles is if you know WHEN the gun was fired. There is NO WAY you can find the radius of the circles you are asking for. If you say the your intersecting radii are 10,000 yards and 11,000 yards.. giving you a 1 second lag between the microphones, All I have to say is what lag would you get at 20,00 and 21,000 yards? The same 1 Second delay. You absolutely need the 3rd microphone. And the distance between the microphones is only needed for drawing the graph.

There may be some algebra or trig that could help you out, but I'm too dumb to figure it out.

I'll try and draw the intersecting circles you are talking about. Tell me exactly the radius of the circles and how far apart they should be and how you figured out how big the radius should be. That would denote distance and I don't see how you figured out WHEN the gun was fired to figure out the distance.

Scott

[Maastrictian]

I have to agree with karch, Jons and others who've posted that two stations is insufficient even under ideal conditions. Unfortunatly, I too am too dumb to do the math

So let me present a challenge to MajorBooBoo. Given the following data, tell us where the firing gun is located.

-Listening stations Alpha and Bravo are 1.4KMs apart. (Alpha is at (0,0). Bravo is at (1400,0))

-Alpha and Bravo are on the front lines, such that all points with a positive Y value are in enemy territory.

-The firing gun must be in enemy territory

-They both hear what they know to be the same gun fire with a time difference of 1 second.

-Alpha hears the sound before Bravo does

-Assume the speed of sound is 350m/s to make thing easier.

Just a possible way of simplifing this dispute.

--Chris

[karch]

based on what Maastrictian said and what maj boobo wants, here is a chart that I think you both are describing.

The big question is how will you ever know the radius of the 2 circles. A-X and B-X?

Scott

[MajorBooBoo]

Originally posted by karch:

based on what Maastrictian said and what maj boobo wants, here is a chart that I think you both are describing.

The big question is how will you ever know the radius of the 2 circles. A-X and B-X?

Scott

The concentric circles should be such that the center of the two circles have X as the center. One circle should have A on the outside of its circum and B should likewise be on the outer circle circum. Busy right now and hope to be home later tonight.

[MajorBooBoo]

Originally posted by Maastrictian:

I have to agree with karch, Jons and others who've posted that two stations is insufficient even under ideal conditions. Unfortunatly, I too am too dumb to do the math

So let me present a challenge to MajorBooBoo. Given the following data, tell us where the firing gun is located.

-Listening stations Alpha and Bravo are 1.4KMs apart. (Alpha is at (0,0). Bravo is at (1400,0))

-Alpha and Bravo are on the front lines, such that all points with a positive Y value are in enemy territory.

-The firing gun must be in enemy territory

-They both hear what they know to be the same gun fire with a time difference of 1 second.

-Alpha hears the sound before Bravo does

-Assume the speed of sound is 350m/s to make thing easier.

Just a possible way of simplifing this dispute.

--Chris

i guess. I am flying my plane right now but have on my calculator watch. the watch has trouble with 1 and 4 (they are on top of each other and give me entry errors ie hit 4 and you get a 14). can I get the 1400 changed to 1300 and 1 sec to two? i will use 650*2 for 1300, I have probs with 7 9sometimes) above the 4 and 1 also. 350 is great.

is this Ok? how will you know I am correct. must fly..strait..

[MajorBooBoo]

difficult conditions but heres answr

slope from origin to gun is prox 2.4, time from gun to origin 1.42 sec, time to (1300,0) is 3.43 sec, gun near (192, 460).

Can anyone confirm?

[MajorBooBoo]

It took me twenty minutes and I am flying a plane. I had to mix some drinks too and theres Tom Collins all over the dash. Hope those guages are working..Anyone need ice? Theres some on the wings...

[karch]

Originally posted by MajorBooBoo:

difficult conditions but heres answr

slope from origin to gun is prox 2.4, time from gun to origin 1.42 sec, time to (1300,0) is 3.43 sec, gun near (192, 460).

I really think that might be ONE solution. But for the sound to arrive at microphone B 1 second after microphone A, there are an infinate number of places for this to happen (well, until the gun is so far away it can't be heard).

Any place where the gun is 340 meters further away from microphone B than A, there will be a one second delay between the microphones picking up the report. I admit I don't have to math to back it up, just my witts.

[ February 15, 2002, 08:23 PM: Message edited by: karch ]

[MajorBooBoo]

Booboo to Nebelwerfer thread..Booboo to Nebelwerfer thread..come in nebelwerfer thread...

Its two seconds karch, I had to change to two seconds? Do the numbers compute? Please check in teh equations

The gun is constrained by the two radii, if you move to the left/right, the radii change and so would the two second time diff, same for other moves. Its constrained...

[JonS]

Originally posted by MajorBooBoo:

Booboo to Nebelwerfer thread..Booboo to Nebelwerfer thread..come in nebelwerfer thread...

Its two seconds karch, I had to change to two seconds? Do the numbers compute? Please check in teh equations

The gun is constrained by the two radii, if you move to the left/right, the radii change and so would the two second time diff, same for other moves. Its constrained...

No need to be offensive MBB.

The circles are not constrained. Rather they are constrained, but only in that one is always 1 second (2 seconds, whatever) larger than the other.

[MajorBooBoo]

Originally posted by JonS:

The circles are not constrained. Rather they are constrained, but only in that one is always 1 second (2 seconds, whatever) larger than the other.

?????????????????

If the distance between the sensors is constant, then moving the noise source in any direction will change the time differential. Accept it or why not try to disprove it? You are just talking in circles.

The only special cases (here we go again) are when the sensors are equidistant or the two sensors and the noise source are in a strait line. What is so hard about this?

Did my calc work? anyone have anything to say besides unsupported gobblygoop?

[ February 15, 2002, 10:38 PM: Message edited by: MajorBooBoo ]

[MajorBooBoo]

Originally posted by karch:

</font><blockquote>quote:</font><hr />Originally posted by MajorBooBoo:

difficult conditions but heres answr

slope from origin to gun is prox 2.4, time from gun to origin 1.42 sec, time to (1300,0) is 3.43 sec, gun near (192, 460).

I really think that might be ONE solution. But for the sound to arrive at microphone B 1 second after microphone A, there are an infinate number of places for this to happen (well, until the gun is so far away it can't be heard).

Any place where the gun is 340 meters further away from microphone B than A, there will be a one second delay between the microphones picking up the report. I admit I don't have to math to back it up, just my witts.</font>

[Michael Dorosh]

Major DooDoo, it IS Friday night. Instead of drinking by your computer (or was that your wrist watch?) and waiting for people to bow to your mathematical mastery, why don't you go out and find, I don't know, a GIRL to talk to?

I'm off to do likewise. We can exchange "Dear Diary Moments" tomorrow morning.

Toodles.

[MajorBooBoo]

Originally posted by Michael Dorosh:

Major DooDoo, it IS Friday night. Instead of drinking by your computer (or was that your wrist watch?) and waiting for people to bow to your mathematical mastery, why don't you go out and find, I don't know, a GIRL to talk to?

I'm off to do likewise. We can exchange "Dear Diary Moments" tomorrow morning.

Toodles.

You think the wife would mind? Its early here still.

You are just going to stand by yourself and drink beer anyway. Dont get your hopes up too high. Thanks for all your great input again.

[karch]

Originally posted by MajorBooBoo:

Well IF there are so MANY, then someone find me some (cant use the mirror value (192, -460) or whatever it is). AND COULD YOU CHECK YOUR WORK? I believe my work checked (pretty close anyway).

Having the argument thats theres infinite solutions but you cant compute any is a clue that something is amiss.

This is pretty tiring , I tell ya.[/QB]

All right, now you are getting condescending. Read this and do my very simple equations. Also, please draw your graph, I don't understand what you are explaining.

ANY POINT 340 METERS FURTHER FROM POINT A THAN FROM POINT B WILL GIVE A ONE SECOND DELAY!!!

Answer these questions given the speed of sound at 340M/S and the 2 listening posts are 340 meters apart.

What will the sound delay be at listening post A if the gun fires...

1) 1 foot from mike B

2) 680 meters from A and 340 Meters from B

3) 1020m from A and 680m from B

4) 1360m from A and 1020m from B

5) 1700m from A and 1360m from B

6) 2040m from A and 1700m from B

7) 10200m from A and 6800m from B

8) 13600m from A and 10200m from B

I could go on forever, or pick any number for B, ad 340 meters and get the same result. Listening post B hears the sound 1 second before listening post A. Every time.

If all you have is two listening posts, it is impossible to do range-finding. I just proved it. You asked for some proof. It's right here. I just showed you 8 possible locations for the gun. There are infinite possibilities because the equation here is:

Ax = Bx + 340

Ax being distance from A to X

Bx being distance from B to X

Any value for Bx + 340 will always give you the same result. A one second delay.

[Yeknodathon]

Dear Carol Vorderman

I'm a very special fan of yer Countdown show. I particularly like Mr Richard Whiteley's ties but I must say, I only watch it because yer nice and so good with letters and, of course, numbers. It makes me day to see you slap those vowels onto the board and do all those mental calculations in your head. Boy, and your legs ain't half bad too.

No, yer see Carol, we're having a spot of bother.

I'm not a number person meself and I was wondering if yer could come down here to add some, well, show-bizz number-crunching savvy to our problem... whatever it is... and I could inspect yer legs... well, forget about the numbers, just bring your legs.

Kind Regards

Yeknod

[MajorBooBoo]

Originally posted by karch:

</font><blockquote>quote:</font><hr />Originally posted by MajorBooBoo:

Well IF there are so MANY, then someone find me some (cant use the mirror value (192, -460) or whatever it is). AND COULD YOU CHECK YOUR WORK? I believe my work checked (pretty close anyway).

.

ANY POINT 340 METERS FURTHER FROM POINT A THAN FROM POINT B WILL GIVE A ONE SECOND DELAY!!!

Answer these questions given the speed of sound at 340M/S and the 2 listening posts are 340 meters apart.

1) 1 foot from mike B

2) 680 meters from A and 340 Meters from B

3) 1020m from A and 680m from B

.

.

.

.[/QB]</font>

[MajorBooBoo]

http://www.carleton.ca/Capital_News/05021999/f2.htm

I dont care about what EVERYBODY says about Canadians, Some of them are pretty smart. They are actually the Finns of the Americas...

Anyone tell me "watt" sound principle they are working on?

[MajorBooBoo]

http://www.tno.nl/instit/fel/museum/en/earlyeq.html

I gotta get me to the Museum Waalsdorp..

Seems they are finding elevation here from two mikes (read the paragraph thouroughly please and dont be misled because theres four mics, two are used for elevation, two are used for angle).

How'd they do it?

[JonS]

Originally posted by MajorBooBoo:

</font><blockquote>quote:</font><hr />Originally posted by karch:

...ANY POINT 340 METERS FURTHER FROM POINT A THAN FROM POINT B WILL GIVE A ONE SECOND DELAY!!!

Answer these questions given the speed of sound at 340M/S and the 2 listening posts are 340 meters apart.

1) 1 foot from mike B

2) 680 meters from A and 340 Meters from B

3) 1020m from A and 680m from B

.

.

.

For the umpteenth time.. you are describing the special case of the two receptors being in line with the noise source. [/QB]</font>