PDA

View Full Version : Math Help

pyroracing85
02-18-2008, 04:34 PM
I need help figuring out how long to make my pieces.

Basically, what I am doing is I have a rectangel that is 48"x36". What I want to do is put two "X"'s in the middle of that rectangle. I figured out the sin and the angle but it is confusing me. How would I put two "X"s in this rectange? I am just trying to figure out the math right now so I can cut my pieces to length and be able to order enough material.

Moz
02-18-2008, 04:47 PM
Do you mean an X in each half?

I would suggest drawing diagonal lines on the work. If you draw the two diagonals on the big rectangle that will give you the centre point so you can divide it into two identical half-sized rectangles. Then draw the diagonals of those and measure them to work out the lengths you want. This also avoids any problems caused by your "rectangle" not being quite square (you can tell that by measuring the diagonals and if the numbers are not all the same... it's not square).

Here's a quick sketch. Start with the red lines to split the outside rectangle in half. Then draw the blue line up the middle through the intersection of the red lines. I've drawn two blue rectangles to make it a little more obvious. Then you can draw the two green crosses that you actually care about. Now measure the four lengths and you're done.

enlpck
02-18-2008, 04:52 PM
If the diagram matches what you want to do, use the pythagorean formula.

If it doesn't, a drawing of what you want will help a lot.

txfireguy2003
02-18-2008, 04:58 PM
http://i230.photobucket.com/albums/ee167/txfireguy2003/rectangle.jpg

Is this what you're trying to do? If so, the length of the diagonal "H" will be (A*A)+(B*B)=(H*H). So for these dimensions, 24" squared + 36" squared = H squared. Take the square root of that number and you get the length of H. For this particular instance, H equals 43.25". That is from the midpoint of the 48" side to the opposite corner. All diagonals will be the same length, but you'll either need to notch them so they will "cross", or cut 2 of them in half and weld to the midpoint of the one that is not cut. Hope that explains it for you, kind of hard to explain in text.

txfireguy2003
02-18-2008, 04:59 PM

denrep
02-18-2008, 05:00 PM
IF:
˝ of 46 x 36 = 24 x 36
Asq + Bsq = Csq
24x24 =576
36x36=1296
576+1296=1872
1872sr = 43.3

Each leg of the X's two legs, will require about 44 inches of stock.

Never mind, I'm too slow, use Moz's or enlpck's system! :waving: Or txfireguy, even :realmad:

Anyone else before I'm done? :laugh:

pyroracing85
02-18-2008, 05:13 PM
If the diagram matches what you want to do, use the pythagorean formula.

If it doesn't, a drawing of what you want will help a lot.

Yes, I did this but that is for a line.. That wouldn't work with a piece of tube that is 2" wide. If I cut them to that length they would hit each other on the ends..

pyroracing85
02-18-2008, 05:18 PM
http://i230.photobucket.com/albums/ee167/txfireguy2003/rectangle.jpg

Is this what you're trying to do? If so, the length of the diagonal "H" will be (A*A)+(B*B)=(H*H). So for these dimensions, 24" squared + 36" squared = H squared. Take the square root of that number and you get the length of H. For this particular instance, H equals 43.25". That is from the midpoint of the 48" side to the opposite corner. All diagonals will be the same length, but you'll either need to notch them so they will "cross", or cut 2 of them in half and weld to the midpoint of the one that is not cut. Hope that explains it for you, kind of hard to explain in text.

but if I did it from the midpoint of 48" the two x's are going to collide with each. That means I would have to do it a little less than the midpoint on either sides.

Moz
02-18-2008, 05:57 PM
You didn't mention the 2" tube bit at the start, so we all left that out. You still need the full length, otherwise when you cut the bevels on the ends you'll be missing a bit at the tips. What you can do is subtract 2" from two of the diagonals to allow for the cut in the middle. I assume you are quoting the inside dimension of the main rectangle? If not, subtract 2" or 4" from the sizes above depending on whether your quoted size is centreline or outside dimension. That will drop 2 3/4" or 5 1/2" off the diagonal length (actually 2.8" but it's close to 2 3/4" in imperial).

I would be tempted to bevel the centreline edges of the diagonals so you have a short weld along the centreline to make it all a little stronger, but also much easier to make (because all the bits are now rotationally symmetric as well as longitudinally mirror symmetric)

denrep
02-18-2008, 06:48 PM
...I am just trying to figure out the math right now so I can cut my pieces to length and be able to order enough material.

Sorry, I was just quoting rough math for the steel order.

txfireguy2003
02-18-2008, 07:06 PM
http://i230.photobucket.com/albums/ee167/txfireguy2003/rectangle-1.jpg

Do it this way. Cut your diagonals square at 38 13/16", then miter them at then ends on a 146 degree angle and a 124 degree angle so that you don't lose any length. Basically, the 38 13/16 length should be at the midpoint of the side, or 1" from either side on both ends. You repeat this precess for the other side of the rectangle so that both "X's" butt up to each other 1.414" from the inside edge of the rectangle. Looks like more trouble than it's worth to me though.