Arch Forms for Oven Door - The Calculation?

[Kemo]

All,

I have been troubled trying to figure out how to make the perfect Door Arch Form for my oven door and landing. I have heard that there are charts and calculators to tell you how to draw out the arch based on rise and run...

Anyway, I was hoping to come up with a DIY method for getting the Door Arch Form, just right, using a string, marker, and a piece of plywood. Basically, i would tack a string in the center of the plywood sheet and run a pen along the outer swing creating an arc. I just need to know how long the string should be.

I would like to make my oven opening one firebrick high at 9" with the width of the opening to be 20". Now, I also want an arch with a max height of 11". What is the radius of the circle I would need to draw on the plywood?

After some heavy cyphering and Internet searching ...

I found some equations from basic geometry online and realized that I have grossly forgotten alot of this stuff over the years. After going over the equations a few times, I threw some numbers at it to see how it all panned out.

Everything seemed to work on paper with this method and I was hoping some of you guys could double check the logic here to make sure I'm not losing it

If it checks out, hopefully it will help someone out with their door arch forms down the road.

I'm just trying to get all this calculated before I start...

[Wlodek]

It might be a stupid question, but does the arch have to be a part of a circle?

I have a slight feeling that an optimal (not sure in what way though) arch has something to do with the curve that a chain makes when you hold it by its ends and let the middle bit droop.

Just to stir things up of course, as it probably does not matter at this scale.



Greetings from sunny Cumbria,

W.

[Kemo]

this is only cosmetic! i was just trying to get a smooth curve given the dimensions. And mostly for sh*ts and giggles

[dmun]

My cad program shows 26 exactly (or to within .001")

[Hendo]

In the calculation where you've arrived at z=y/2sin(x), you've assumed that you're working with a right-angled triangle, but I don't think it is. The angle appears to be slightly less than 90°, and would reduce with increasing opening width.

This might explain the small discrepancy between David's CAD figure and your calculated one.

Paul.

[Dutchoven]

The answer is no, it does not have to. The shape you refer to is a catenary arc which is the arc created when a chain of equal density is hung and gravity is allowed to act upon it uniformly. It is very stable physically. In the case of a door arch it would be called an inverted catenary. For brick arch forms it is a bit simpler to do the arc of a circle I think.
Best
Dutch

Originally Posted by Wlodek

It might be a stupid question, but does the arch have to be a part of a circle?

I have a slight feeling that an optimal (not sure in what way though) arch has something to do with the curve that a chain makes when you hold it by its ends and let the middle bit droop.

Just to stir things up of course, as it probably does not matter at this scale.



Greetings from sunny Cumbria,

W.

[Wlodek]

Just out of interest I calculated and plotted the 3 arches (circular, parabolic and simple catenary) for a specific height/width ratio of the opening of 1/2 and the plot is attached.

Interestingly, for this particular ratio, the catenary is very close to the parabolic one. Presumably meaning that not only do you get the focal effect, but also the strength of the catenary for pretty much the same arch .... not that it matters with the door.

Oh, dear, do I have too much time? No!

W.