…and yes, I googled it. But since I suck at math and don’t know what I don’t know, or the correct terms in which to describe my question, I couldn’t find it, or figure out how to use what I did find to answer my question.
Okay so, I’ve got a pass-through in my kitchen and I want to put up an arch in it. I need to figure out how to figure out how long the piece of material needs to be. I want to use some form of veneer or other very thin wood type material so that it makes a nice arc shape. I want the top of the arch (of course) to touch at the center of the pass-through.
The counter top is 92 and 7/8th inches long and the height of the pass-through is 49 and 7/8ths inches. Not that that matters, but if one of our doper mathletes could provide me with an equation I would be EVER so grateful.
One way to do this would be to make half an ellipse with a major axis of 92+7/8 inches and a minor axis of 49+7/8 inches.
Finding the circumference of an ellipse actually requires a bit of fancy calculus, but we can plug the numbers into the formula with [+%29+%2F+2"]Wolfram Alpha](( 4*(92.875)* EllipticE - Wolfram|Alpha[1±+%2849.875%29^2%2F%2892.875%29^2) and divide it in half, getting a length of roughly 229.347 inches. I think that’s right.
How far down do you want it to come? A few inches? A foot? All the way to the counter? Do you want it to be circular or elliptical?
It might be easiest to get something thin and flexible (like a piece of vaneer or formica) that’s about 10 or so feet long and just hold it up in front of the area you want to make the arch. If you had some friends help you it would make this job a lot easier. Have one person stand on the other side and hold the center of your materiel to the center of the pass through, then have another (or two) friend bend the sides down. When it get’s to where you want it mark the spot where it crosses the edges of the pass through and measure it.
I just added the length and height and divided by three to get an approximate average for the radius of a perfectly circular arch and I get 149 1/2 inches.
Friedo messed up in using 98 inches instead of 98/2 as one of the axis of his ellipse and consequently his figure is way too large. As a gut check, it is longer than the three sides the arch will fall inside of.
The larger number. I can always cut my piece of veneer shorter if need be! I also got a second in person bit of help at work. 229 inches just SOUNDS right, I have a gut (decorator’s?) instinct of what looks right, just no innate math understanding on how to get there. I’ll probably work the equations provided a few times to make sure I can arrive at the same answer three times.
