I am tasked, by my fiance, to build a “diamond” bookshelf, a la this (she saw it in a commercial).
The space is roughly 6’6" across, and up to 8 feet high (the height of the ceiling, but it needn’t be full floor-to-ceiling).
If you can’t view the linked picture for whatever reason, it’s a bookshelf consisting of a frame filled with square cubbies for the books, but at a 45 degree angle, so the cubbies are diamonds.
Anyway, if I want the diamonds to be 1.5 feet square, how do I calculate the length for the diagonal boards? Surely there is a way to figure this out without building a trial-and-error mock-up, but I’m brain-farting on how.
Unless I’m misunderstanding you, this is just a pythagorean theorem problem. Your answer is the square root of (1.5^2 + 1.5^2). In other words, 2.123 feet.
Surely if you want the diamonds to be 1.5 feet square then the length of the sides is… 1.5 feet!
The “diamonds” are just squares, but tilted through 45 degrees. That doesn’t alter the length of their sides.
Or do you mean you want the vertical height of each diamond to be 1.5 feet? In that case, 1.5 feet is the diagonal of the square, or in other words the hypotenuse of a right-angled isosceles triangle. Therefore the sides of the square would be 18 inches x (sqrt2)/2 or 12.73 inches, roughly.
No, it would be 2.25 square feet, which is the same as 1.5 feet square. The OP said “1.5 feet square”, which means 1.5ft on each side. The same as a 10ft by 10ft room is 10 feet square, or 100 square feet. Terminology.
Anyway, assuming the side length is to be 18 inches, then each cubby will be about 25.46 inches high vertically.
So in an 8ft high room you could fit three complete cubbies high. In which case each long board needs to be 6 x 18 = 108 inches long, as it will form the sides of 6 cubbies. (Look at the picture and imagine it cut off to exactly three cubbies high - each diagonal board will form five complete cubbies plus a half-cubby at the top or bottom.)
That is ignoring the thickness of the boards in the calculations, of course.
Square root of (the length squared + the height squared. In the case of 45 degree angles the length of the diagonal will be 1.4142 x length of the side. In your case it’s a little more complicated because diagonals are cut off and what not.
You want the lozenges to be 1.5 ft square (not 1.5 sq ft which is different). With each side 18" the diagonals of the individual lozenges are just over 25.45 inches. This translates into approximately 3.065 lozenges across at 78". If you were to make it exactly 3 across, then <insert math> the sides of each lozenge would just under 18.4" inches and if it is the outside dimension that is 6’6", then the lozenges would be even smaller making them closer to your desired size.
But even if you don’t do it that way the math is easy. If the diagonal is going from top (or bottom) to a side, look at it like a triangle and the diagonal is sqrt(2) x distance from the corner along either side. if it goes from top to bottom it is (inside height) x sqrt(2). From side to side it would be (inside width) x sqrt(2).
ETA: This is assuming 1 board goes side to side of the cabinet forming the sides of all of the lozenges.
No, the sides of the diamonds will be 1.5 feet, but the diamonds will be made up of long interlocking boards inside a frame at a 45 degree angle, so I need the length of the whole board across the frame (a single board will be the side of several diamonds).
See my post number 8. If you only want a whole number of diamonds, then it will be three diamonds high, roughly 76 inches. Each long diagonal board will be six “squares” long, or 108 inches.
After playing around with the numbers, I came up with a frame 6’4" across and 7’5" high, and diagonal boards, from the bottom up, of 1’6", 4’6", 7’6", 9’, 6’, and 3’ going both ways, which gives 15 whole squares, 11 halves, and 2 quarters.
My sketch here on photobucket. Please tell me it looks right.
Jeez, guys, I have NEVER seen a simple problem become so complex in such a short time. I mean, really, anyone who’s taken the first week of Plane Geometry can figure this out in 10 seconds max. Are you all brain-dead from too much Halloween candy?
Without sketching it all out properly I can’t be sure of the details, but I think the thickness of the wood could easily cause some complications, with so many “crossings”. You’ll need to make sure that your measurements take account of it, otherwise your boxes won’t be quite square and the whole thing will look a bit off. The simplest way would probably be to plan the dimensions using the centre line of the wood, and then reduce the board lengths by half of the thickness of the wood where applicable.
