MATH HELP (Geometry)

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SDA1C
SDA1C Posts: 2,069
edited September 2013 in The Clubhouse
Hi Folks,

I desperately need formula assistance with a 3d model. I need to transition from a 12 ft by 14 ft rectangle hopper to a 1 ft by 1ft square. It is for a grain load out tank. The angle has to remain as close to 45 degrees as possible but a little leeway is afforded. I need to be able to plug in the height and get the cut angles for the sheets. We do not have Solidworks or the like as of yet. Does anyone know if there is a formula for pyramids of unequal sides that will give me the congruent angles and sides given the from/to/elevation dimensions? :eek:

Thanks, 1c
Too much **** to list....
Post edited by SDA1C on
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  • exalted512
    exalted512 Posts: 10,735
    edited September 2013
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    I don't understand what you're wanting exactly...can you draw a picture?
    Music is like candy, you have to get rid of the rappers to enjoy it
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  • BlueFox
    BlueFox Posts: 15,251
    edited September 2013
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    SDA1C wrote: »
    Does anyone know if there is a formula for pyramids of unequal sides that will give me the congruent angles and sides given the from/to/elevation dimensions?

    The angle of the dangle is equal to the mass of the a$$ so long as the thrust of the bust remains constant. Or something like that. :redface:
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  • Msabot1
    Msabot1 Posts: 2,098
    edited September 2013
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    Maybe I'm missing something here....how can you have a square with 45 degree angles?
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  • SDA1C
    SDA1C Posts: 2,069
    edited September 2013
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    Sheet cut formula.doc Its a chicken scratch but a place to start
    Too much **** to list....
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  • SDA1C
    SDA1C Posts: 2,069
    edited September 2013
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    Imagine an upside down pyramid with two sides at 14 ft and two sides at 12 feet, then truncated at 1 ft by 1 ft opening at the bottom. The sides of the pyramid are 45 to the ground as they taper to the small end.
    Too much **** to list....
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  • Msabot1
    Msabot1 Posts: 2,098
    edited September 2013
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    Gotcha...realized you were talking downward slope the square is the feed right?
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  • cincycat13
    cincycat13 Posts: 882
    edited September 2013
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    You can use sin, cos, tan and Excel to model it in 2D triangles to make it work if I am visualizing correctly. I am assuming your "45 degree angle" is your slope of fall. Do you want the 12' side to slope 45 floating the 14 tolerance for vice versa?
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  • SDA1C
    SDA1C Posts: 2,069
    edited September 2013
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    Cincy,

    Either is acceptable. Mostly I need a formula to get the cut angle to end up with the proper side angle when assembled. i am fairly certain the congruent sides need to be...well...congruent. That tells me the angle will be different for the 12 vs. the 14 correct?
    Too much **** to list....
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  • cincycat13
    cincycat13 Posts: 882
    edited September 2013
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    shoot me a pm with an email addy and I will work on it tonight if you are not in a hurry and have excel access. remember excel does trig in radians not degrees.

    Basically you need to drop the 12' side 7' in a 7' run to maintain 45 which gives you a 7 x 7 x 9.9' hypotenuse using a^2 + b^2 = c^2.

    Then the 14' side must drop 7' in a 6' run in order to match up. this yields a 9.22' hypotenuse. using ATAN 7/6 gives you a 49.3 degree of slope for that side.

    now that all dimensions are fixed, you can use trig sin/cos/tan to get the angles. Then use the laws of similar triangles to shorten up the 9.90 and 9.22 dimensions to yield your 1 x 1 opening instead of a point.
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  • cincycat13
    cincycat13 Posts: 882
    edited September 2013
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    i'd check the math or build to scale from cardboard, but here were my numbers...
    bin.pdf 131.2K
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  • JimAckley
    JimAckley Posts: 1,138
    edited September 2013
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    DSkip is a champ at math if you still need any help.

    http://www.polkaudio.com/forums/member.php?60503-DSkip
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  • SDA1C
    SDA1C Posts: 2,069
    edited September 2013
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    I think I got it with the subtraction of the trunk. I went to an 11 by 13 6 ft high pyramid and went from there. After the final pyramid is identified I can just stretch the side back out a foot...I hope.
    Too much **** to list....
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  • deronb1
    deronb1 Posts: 5,021
    edited September 2013
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    It sounds like you need to find the height of the triangles(ish). The height of the long side would be 72 in and the length of the short side would be 84 in. Of course you would be working with rectangles, but it seems you could cut them at a 45 and then cut out the ends to make your 1 x 1. Not sure if that would work or not.
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  • maximillian
    maximillian Posts: 2,142
    edited September 2013
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    I calculated the height to be 12.5 feet. Unfortunately, I have some other work so I can't confirm it in something like Solidworks. However, you can calculate a diagonal line (call it I) running along the top from the corner of the top to the corner of an imaginary 1'x1' box in the center of the top rectangle:

    I = sqrt (6.5^2 + 5.5^2) = 8.51 feet.

    The diagonal edge (D) from the top corner to the bottom corner (along the outer edge) is formed by an isosceles triangle with angle 45° and edge equal to:
    (14-1)/2 = 6.5'. So the diagonal (D) is: 6.5 / cos(45) = 9.19

    So now you have two sides of a right triangle where the third is the height. The triangle is a diagonal edge of the cone running top to bottom. So the height of your cone is:
    h = sqrt(9.19^2 + 8.51^2) = 12.52

    Please someone confirm before you cut twice. It's been a long day.

    If the above is correct, you can plug them into Excel. You can also back track the formulas so that given the height you can calculate the angle (used 45 above).
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  • cincycat13
    cincycat13 Posts: 882
    edited September 2013
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    Max, not sure the cone could be over 12' if one side of the rectangle is 12'. I figured the height at 7' to allow for a 45 slope

    I did some math and art with my 6yr old this am and folded to the dimensions I posted in my PDF. I had to shoot it on the MicroPro 3000 for scale as this is an audio forum:wink:
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  • maximillian
    maximillian Posts: 2,142
    edited September 2013
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    Hey nice approach, especially involving your kid.

    Solidworks says the height should be 8.5' for a 45° edge slow with respect to the top face. Looks like I was wrong with my chicken scratch. Sry.

    pyramid.jpg
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  • SDA1C
    SDA1C Posts: 2,069
    edited September 2013
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    Thanks for the input folks. The new perspectives have really given me many new angle of approach that I previously didn't explore. I went with a concept derived from the combination of a few different points of view and settled on this...

    Reduce the trapezoids to triangles by eliminating the 12"x12" opening and bringing two sides to 11 ft and two sides to 13 feet making a rectangle pyramid and omitting the truncation. I then found the congruent leg length by Pythagorean theory. I used the leg length to identify the two common legs and found angles by the two leg lengths of 10'4" and the given leg lengths of 11 and 13. Once the angles were identified I added a foot back into the triangle making it a trapezoid once again with a 1 ft by 12/14 ft parallel sides.
    Too much **** to list....
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  • maximillian
    maximillian Posts: 2,142
    edited September 2013
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    Glad you got it!

    I realize my mistake above. Not sure why I made it more complex than it needs to be. An imaginary diagonal line along the top surface connecting the outer corner to a 1x1 corner is like I mentioned above, 8.51'. Then since you want the diagonal of the top corner to the bottom corner to be 45°, then the height is simply:

    H = 8.51 * tan (45°) = 8.51. This is what Solidworks gave me. From there you can calculate the diagonal length:

    DL = 8.51 / cos(45) = 12.03.

    So the sheet metal corner angle along the 14' side is:
    cos-1(Θ) = 6.5/12.03, so Θ is 53°.

    Both these numbers are confirmed by Solidworks.
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  • SDA1C
    SDA1C Posts: 2,069
    edited September 2013
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    I am afraid this wont show but I am off for the weekend and the actual paper is on my desk. The angles if I remember correctly were 77/51/52 and 58/58/64 with odd leg being 13 and 11 respectively. They all share 2 legs at 10'4 or so...I'll do the exacts again Wednesday after this 4 day weekend that was strategically, I mean coincidentally:cool:, the weekend after the close of the last order.

    *The scan isn't worth posting*
    Too much **** to list....
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  • scottyboy76
    scottyboy76 Posts: 2,905
    edited September 2013
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    Yeah, but nonoe of you guys figured in gravity.
    humpty dumpty was pushed
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