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Math Help - Question 7

  1. #1
    Grand Panjandrum
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    Question 7

    I hope this is an easy one

    A thin semi-infinite rod is made of high explosive. The detonation speed is v
    (that is the speed at which the detonation propagates down the rod is v)
    The speed of the detonation products through space is u<v.

    Describe the shape of the shape of the explosion in time and space. That is
    what is the time evolution of the region occupied by the explosion products.

    RonL
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  2. #2
    Super Member malaygoel's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    I hope this is an easy one

    A thin semi-infinite rod is made of high explosive. The detonation speed is v
    (that is the speed at which the detonation propagates down the rod is v)
    The speed of the detonation products through space is u<v.

    Describe the shape of the shape of the explosion in time and space. That is
    what is the time evolution of the region occupied by the explosion products.

    RonL
    Is there any specific direction in which detonation produvts would travel?

    Keep Smiling
    Malay
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by malaygoel View Post
    Is there any specific direction in which detonation produvts would travel?

    Keep Smiling
    Malay
    Unlike in SF movies the products from a small amount of explosive will expand
    in a shperical shell.

    RonL
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  4. #4
    Super Member malaygoel's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    Unlike in SF movies the products from a small amount of explosive will expand
    in a shperical shell.

    RonL
    what would we the centre of the spherical shell?
    Thanks for your response

    Keep Smiling
    Malay

    The point of detonation of the small quantity of explosive.

    RonL
    Last edited by CaptainBlack; November 21st 2006 at 09:28 AM.
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  5. #5
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    Would it be a solid cone that increases in base radius at the speed of the explosion products and hiegth at the speed of the explosion's movment up the rod? At the base of the cone, there is a semisphere whose radius increases with the explosion products' speed as well.
    The cone angle is 90u / v˚, since slope can be found by rise (u) over run (v) and be compared to the angle of the x- and y-axes (the y-axis is the cone's hypotenuse). It will always be less than 90˚, since the speed of the explosion particles is less than that of the explosion itself.
    Last edited by The Pondermatic; November 21st 2006 at 02:10 PM.
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  6. #6
    MHF Contributor Quick's Avatar
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    Sorry, your Web browser is not Java compatible. This prototype will not be of much interest to you.

    Move the green point to see the explosion









    The speed of the detonation products through space is set by the distance from S to U (which you can change)
    Last edited by Quick; November 21st 2006 at 02:56 PM.
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  7. #7
    Grand Panjandrum
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    Quote Originally Posted by Quick View Post
    [Move the green point to see the explosion









    The speed of the detonation products through space is set by the distance from S to U (which you can change)
    And what is the angle involved?

    RonL
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  8. #8
    MHF Contributor Quick's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    And what is the angle involved?

    RonL
    What angle?
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  9. #9
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Quick View Post
    What angle?
    The angle the "cone" makes when projected against a flat surface. ie. what is the angle the top of the cone makes?

    -Dan
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  10. #10
    Grand Panjandrum
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    Solution

    First ask what happens when a single, tiny particle explodes. It will explode in all directions, giving a spherical explosion. The radius of the sphere containing the explosion products will expand at speed u.

    Now fir the semi-infinite rod, consider it to be made up chain of particles of the explosive. Each particle generates a spherical explosion that spreads through space with speed u. However, the particles do not all explode at the same time. The explosion propagates down the rod with speed v. The result is a series of spheres, the largest growing from the end of the rod and having radius ut, and as you proceed down the rod to later explosions, successively smaller spheres, until you arrive at the point where the explosion is just occurring. The successively shrinking spheres add up, as shown in the attached figure, to a cone of altitude vt with a hemispherical cap of radius ut fitted perfectly on to the base of the cone.

    RonL
    Attached Thumbnails Attached Thumbnails Question 7-gash.jpg  
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  11. #11
    Senior Member TriKri's Avatar
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    A truly great illustration. That's pedagogy.
    It reminds me of a sound barrier from a plane going in over the speed of sound.
    And a really nice illustration from Quick as well.
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  12. #12
    MHF Contributor Quick's Avatar
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    Quote Originally Posted by TriKri View Post
    And a really nice illustration from Quick as well.
    I agree. Too bad MathGuru is considering dropping the program from the forum

    To get that diagram, type in:
    [draw]FixedPoint (130,50); FixedPoint (130,250); Ray (2,1) [hidden]; Point on object (3,.05) [green]; Point on object (3,0) [hidden]; Point on object (3,.9) [hidden]; Point on object (3,.8) [hidden]; Point on object (3,.7) [hidden]; Point on object (3,.6) [hidden]; Point on object (3,.5) [hidden]; Point on object (3,.4) [hidden]; Point on object (3,.3) [hidden]; Point on object (3,.2) [hidden]; Point on object (3,.1) [hidden]; Point on object (3,1) [hidden]; Segment (1,2) [thick]; Segment (1,4)[hidden]; Length (17,0,0,'d') [hidden]; FixedPoint (20,150) [label ('S')]; FixedPoint (20,250) [label ('V')]; Segment (19,20); Point on object (21,.5) [label ('U')]; Segment (19,22); Length (21,0,0,'d') [hidden]; Length (23,0,0,'25') [hidden]; Calculate (0,0,'26','A B /') (25,24) [hidden]; Calculate (0,0,'0','A 20 - B *') (18,26) [hidden]; Calculate (0,0,'0','A 40 - B *') (18,26) [hidden]; Calculate (0,0,'0','A 60 - B *') (18,26) [hidden]; Calculate (0,0,'0','A 80 - B *') (18,26) [hidden]; Calculate (0,0,'0','A 100 - B *') (18,26) [hidden]; Calculate (0,0,'0','A 120 - B *') (18,26) [hidden]; Calculate (0,0,'0','A 140 - B *') (18,26) [hidden]; Calculate (0,0,'0','A 160 - B *') (18,26) [hidden]; Calculate (0,0,'0','A 180 - B *') (18,26) [hidden]; Calculate (0,0,'36','A 200 - B *') (18,26) [hidden]; Calculate (0,0,'37','A B *') (18,26); Circle by radius (1,37) [hidden]; Circle by radius (2,36) [hidden]; Circle by radius (6,35) [hidden]; Circle by radius (7,34) [hidden]; Circle by radius (8,33) [hidden]; Circle by radius (9,32) [hidden]; Circle by radius (10,31) [hidden]; Circle by radius (11,30) [hidden]; Circle by radius (12,29) [hidden]; Circle by radius (13,28) [hidden]; Circle by radius (14,27) [hidden]; Circle interior (38) [red]; Circle interior (39) [red]; Circle interior (40) [red]; Circle interior (41) [red]; Circle interior (42) [red]; Circle interior (43) [red]; Circle interior (44) [red]; Circle interior (45) [red]; Circle interior (46) [red]; Circle interior (47) [red]; Circle interior (48) [red];[/draw]
    Last edited by Quick; November 27th 2006 at 01:37 PM.
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  13. #13
    Grand Panjandrum
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    Quote Originally Posted by TriKri View Post
    A truly great illustration. That's pedagogy.
    It reminds me of a sound barrier from a plane going in over the speed of sound.
    It should do, its supposed to illustrate the formation of a Mach cone.

    RonL
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