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  #11  
Old 01-27-2012, 03:26 PM
david s's Avatar
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Default Re: Arch Forms for Oven Door - The Calculation?

There is an advantage in having the sides of the arch steeper than the centre. This becomes self evident when you begin to work the oven. This set up is actually opposite to what you get with a catenary arch. It is easy to sketch the curve on paper, then fold it in half to attain a symmetrical shape.
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  #12  
Old 01-27-2012, 04:46 PM
Serf
 
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Default Re: Arch Forms for Oven Door - The Calculation?

Laku:
Yes, you are of course right.

Z = 24 and as R = Z + 2, R = 26

As an excuse I maintain that the late night online poker game also involved a bottle or two of red wine, must have got a bit distracted by the game, or the glass, toward the end of that calculation.

I'm amazed that a four year old thread gets so many immediate responses. I llok forward to this sort of interest when I get around to building my own... Soon...
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  #13  
Old 01-28-2012, 01:09 PM
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Default Re: Arch Forms for Oven Door - The Calculation?

Use an ellipse.

Use two nails, a piece of string and a pencil:

http://www.mathopenref.com/constellipse1.html

That's it, done. No math required.

Last edited by Neil2; 01-28-2012 at 01:21 PM.
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  #14  
Old 01-28-2012, 04:51 PM
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Default Re: Arch Forms for Oven Door - The Calculation?

Why all this fear of a little mathematics?

String & pencils may work for some, but the mathematical method goes somewhat further in answering the original potser's question.
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  #15  
Old 01-29-2012, 10:16 AM
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Default Re: Arch Forms for Oven Door - The Calculation?

From Steve's original post:

"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."

The ancient Roman and the Medieval/Renaissance stone builders did not have computers.

The probably used strings, nails and pencils.

The Advantages Of The Elliptical Arch | LIVESTRONG.COM

Last edited by Neil2; 01-29-2012 at 10:28 AM.
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  #16  
Old 01-29-2012, 11:09 AM
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Default Re: Arch Forms for Oven Door - The Calculation?

I use a plastic strip or aluminium ruler or a thin piece of plywood, bend to shape with the aid of a few nails and mark.
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  #17  
Old 01-30-2012, 05:47 PM
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Default Re: Arch Forms for Oven Door - The Calculation?

Neil2:
The paragraph you quote from goes on to say "I just need to know how long the string should be." Another paragraph contains "What is the radius of the circle I would need to draw on the plywood? "

This indicates the original poster was happy to use string and pencil to draw the arch, but wanted to know how large a radius was required for his requirements. The calculated method would give him exactly that.
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  #18  
Old 03-15-2012, 09:20 AM
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Default Re: Arch Forms for Oven Door - The Calculation?

Quote:
Saves all that bother wih the nightmares of highschool trigonometry!
Very nice and elegant solution. Not only does it save bother with trigonometry but it is also more accurate:

Since nobody took the trouble to correct kemos original solution here is my go!
I'll use powindahs notations for the lines:

kemos original equations 1 and 2 are OK albeit with rounding errors becuase of the irrational numbers involved - eg sqrt of 104 etc. But I have no idea where he came up with equation 3.

anyway here goes.
Angle X = 11.31
Lets call A the angle between Y and R
and B the angle between Y and Z
and C the angle between G and R
A = B = 180-(90+11.31) = 78.69
C = 78.69 - 11.31 = 67.38
C is an angle of the right angled triangle hypoteneuse R and adjacent G,
cosine C = G/R = 10/R = .384
R = 10/cosC = 10/.384 = 26.0416667
and of course R=Z
more accurate but still not as good as powindahs solution which is exactly right
This thread btw is only for people with too much time on their hands
Amac
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  #19  
Old 03-16-2012, 02:41 AM
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Default Re: Arch Forms for Oven Door - The Calculation?

I wouldn't dispute that too much wotavidone - just I do like to see puzzle solved in an elegant manner, and powindahs was elegant compared with the trigonometric calculations.
One thing though. Boat builders are probably trying to achieve a different curve than arch builders and will be limited by the strain that the lath will sustain without breaking. I would say it would be good for a gentle curve but for anything more rounded it is surely easier to work out the radius and use a compass to draw the outline. Certainly in my case since I made a completely semicircular arch and all I needed to know was what entry width I could tolerate.
Aidan
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  #20  
Old 03-16-2012, 09:03 AM
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Default Re: Arch Forms for Oven Door - The Calculation?

Quote:
no slur upon your trigonometric genius was intended
if there is any ingenuity involved I can asuure you it isn't mine - maybe powindah if he's still around might take umbrage.
I was looking at the original problem and powindahs solution and it struck me how coincidental that it should be an integer. It seemed to me to be highly unlikely but it is correct.
A side effect of this is that the right triangle whose hypoteneuse is the radius G (26) and also the other two sides (10 and 24) are an example of a pythagorean triple - i.e a right triangle whose sides are all integers.

In the list from wiki the second one in bold 5:12:13 (10:24:26) is the relevant ratio for this problem.

Quote:
Common Pythagorean triples
The following are all the Pythagorean triple ratios expressed in lowest form with both non-hypotenuse sides less than 100:

3: 4 :5
5: 12 :13
8: 15 :17
7: 24 :25
9: 40 :41
11: 60 :61
12: 35 :37
13: 84 :85
Another interesting thing (I really do have too much time on my hands) is the relationship of these "triples" to fibonacci series.
Wiki again:
Quote:
Starting with 5, every other Fibonacci number {0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ...} is the length of the hypotenuse of a right triangle with integral sides, or in other words, the largest number in a Pythagorean triple. The length of the longer leg of this triangle is equal to the sum of the three sides of the preceding triangle in this series of triangles, and the shorter leg is equal to the difference between the preceding bypassed Fibonacci number and the shorter leg of the preceding triangle.
So in our case 13 is the hypoteneuse, the long leg is 3+4+5 (12) and the short leg = 8-3 (5)
This is all highly coincidental to the original problem but how cool is it that your arch is based on pythagorean triples, connected to fibonacci series which in turn has connections to the golden ratio 1.61801 which the fibonacci series approaches
I really must have a look to see how kemos oven turned out with all those magical properties built in
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