# Need layout / math help...please



## WARD (Nov 21, 2006)

I need some help trying layout a wall. I'll see if I can explain.

I have a wall 160' long. At 35'8" the top starts a radius. The length of the radius is 181' 11". I am trying to find a way to layout the top of the wall. My plan is to use a chord across the circle and measure up to the top of the wall. I would move horizontally across the chord in one foot increments, and measure up at a right angle to get my wall height. Problem is I can't figure out how to get those measurements from the chord up to the perimeter, (circumference) / top of the wall, from the chord.

I might be going about this all the wrong way. If I am how do I do it???

Trying to attach a chicken scratch drawing to help. I need to know how to find the length of the blue lines.


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## Gus Dering (Oct 14, 2008)

I'm not sure about each one foot increment's height.

The height at the center of the radius (segment height) is about 18' 7 51/64", give or take. :jester:

Here is a link to a decent circle calculator. 
http://www.1728.com/circsect.htm

I hope someone can shed some light on how to figure those incremental heights.

Cutting a pattern surely poses a few challenges.:laughing:


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## lukachuki (Feb 11, 2005)

WARD said:


> I need some help trying layout a wall. I'll see if I can explain.
> 
> I have a wall 160' long. At 35'8" the top starts a radius. The length of the radius is 181' 11". I am trying to find a way to layout the top of the wall. My plan is to use a chord across the circle and measure up to the top of the wall. I would move horizontally across the chord in one foot increments, and measure up at a right angle to get my wall height. Problem is I can't figure out how to get those measurements from the chord up to the perimeter, (circumference) / top of the wall, from the chord.
> 
> ...


Good question and tough problem....I don't have a good answer but I bet someone will jump on it.


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## Tscarborough (Feb 25, 2006)

http://en.wikipedia.org/wiki/Pythagorean_theorem


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## lukachuki (Feb 11, 2005)

Tscarborough said:


> http://en.wikipedia.org/wiki/Pythagorean_theorem


You mean he had more to say than a2 + b2 = c2 Incidentally about the most useful thing i learned in HS.


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## ScipioAfricanus (Sep 13, 2008)

Hopefully this will help.

This is why I love computers and cad programs. :clap:

Andy.


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## stuart45 (Oct 7, 2009)

I think he wants the 181ft 11 ins as the radius, rather than the diameter which would make the top much lower.


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## lukachuki (Feb 11, 2005)

Ok duh. I just threw it on sketchup google's free cad program. Note that the measurments are every 1' 3" As you can figure out I started from the middle and worked my way to the outside of the circle. To go the other way use the same measurements.

~ 18' 6 11/16"
~ 18' 6 1/2"
~ 18' 6 1/4"
~ 18' 6"
~ 18' 5 3/4"
~ 18' 5 5/16"
~ 18' 4 5/8"
~ 18' 3 15/16"
~ 18' 3 3/16"
~ 18' 2 1/2"
~ 18' 1 3/8"
~ 18' 3/16"
~ 17' 11"
~ 17' 9 7/8"
~ 17' 8 1/2"
~ 17' 6 7/8"
~ 17' 5 3/16"
~ 17' 3 9/16"
~ 17' 1 7/8"
~ 16' 11 7/8"
~ 16' 9 3/4"
~ 16' 7 5/8"
~ 16' 5 1/2"
~ 16' 3 1/4"
~ 16' 5/8"
~ 15' 10"
~15' 7 3/8"
~ 15' 4 3/4"
~ 15' 1 13/16"
~ 14' 10 11/16"
~ 14' 7 9/16"
~ 14' 4 7/16"
~ 14' 1 1/4"
~ 13' 9 5/8"
~ 13' 6"
~ 13' 2 7/16"
~ 12' 6 13/16"
~ 12' 2 11/16"
~ 11' 10 5/8"
~ 11' 6 1/2"
~ 11' 2 3/16"
~ 10' 9 9/16"
~ 10' 5"
~ 10' 3/8"
~ 9' 7 3/4"
~ 9' 2 5/8"
~ 8' 9 7/16"
~ 8' 4 5/16"
~7' 11 3/16"
~ 7' 5 11/16"
~ 7'
~ 6' 6 5/16"
~ 6' 11/16"
~ 5' 6 3/4"
~ 5' 9/16"
~ 4' 6 3/8"
~ 4' 1/8"
~ 3' 5 13/16"
~ 2' 11 1/16"
~ 2' 4 1/4"
~ 1' 9 1/2"
~ 1' 2 11/16"
~ 7 3/8"


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## ScipioAfricanus (Sep 13, 2008)

> I think he wants the 181ft 11 ins as the radius, rather than the diameter which would make the top much lower.


In the words of Homer Simpson...

Doh!!!!

Andy.


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## LadyGC (Jul 8, 2010)

Ward: 

Just so I'm clear: with a 181'11" radius under an arch - you are building a 363'10" wall.


You guys are making this harder than it really is. Here's an architectural math trick. (If I sound condascending - I'm really not, I just have a much better way to get to the point than the geeky math whizzes.)

On your chicken scratch: throw your chord out

Pretend your circle is a clock. The radius is 181' 11"

Starting from the centerpoint - AND only looking a 1/4 of the clock

Think of 12:00 as 181' 11" going right to 3:00 ticks down to zero 

Again 12:00 is 181'11" going counterclockwise to 9:00 ticks down to zero.


How ever many increments you are busting up that 1/4 pie is what you divide by. **Remember that's to the centerpoint - we threw out your chord.


Your's is easy - you're going in 1 ft increments. Just count down .. Centerpoint is 181'11" then 180'11, then 179"11 etc (Basically for every measurement you move over you the take the same off your centerline.)


*Example 2:* Lets pretend your chicken scratch is our print for 7 evenly spaced columns under an arch. Radius is 181'11" This time lets keep your chord - 35'8" from top center. That makes our centerpoint (from your chord) now 35'8" 
(Your chord should be 146'3" up... .. 146'3" + 35"8" =181'11") 


Notice we broke that 1/4 pie into 4 pieces. 


That column smack in the middle is our centerpoint - It's 35'8" Going right from center the next column is 3/4 of 35'8" Next is 2/4 of
35'8" and last column on right is 1/4 of 35'8"



Starting on the left that first column is 1/4 of 35'8" 2nd column 2/4, third column 3/4 of 35'8" back to center column (35'8")


*Example 3:* If you really have your heart set on your chord - you turned it into rocket science because you changed the radial centerpoint.


We already determined from example 2 that your centerline is 35'8" Your chord is 160' From centerpoint to end is 80'
Since you are going in 1 foot increments that's 80 increments. 
centerpoint is 35'8" or (428 inches) from top going right next foot is 79/80 of 428, next is 78/80 of 428, etc. 


formula would be 428 times 1/80 divided by 12 = 5.35 foot (converted from decimal is 5' 4 1/5") for your first increment.



CAD is not hard to learn and is awesome if you're doing your own layouts. Most universities offer a 2-3 week class under their continuing education in the summer at a reasonable price. (You then get CAD free for 2 years as a student) BIM is a different story. That's a pretty tough program to grasp. (as for me anyway)


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## WARD (Nov 21, 2006)

Thanks for the help. 
I figured it out last night while playing cards. I deffinately made it too hard. Pythagorean Theorem was right. It was much easier once I got rid of the chord I was trying to measure from. 

CAD is great, but that is part of my problem. I rely too much on computers and calculators. Now when I actual have to use my head, well...:blink:

Anyway, thanks again.


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## wolf1728 (Jun 18, 2011)

Sorry for "bumping" a year-old thread but the address to the "decent circle calculator" (as Gus Dering called it) is: http://www.1728.org/circsect.htm

(The domain name 1728.com was recently stolen).


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## greg24k (May 19, 2007)

Are you building a coliseum? That is one big a$$ circle


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## Paul B (Mar 10, 2007)

LadyGC said:


> Ward:
> 
> Just so I'm clear: with a 181'11" radius under an arch - you are building a 363'10" wall.
> 
> ...


I hope your joking. Are you saying that the thing is a 45 degree slope, one inch down for each inch from center?


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## wolf1728 (Jun 18, 2011)

Well, I don't mean to complicate this even more, but I redrew the very first diagram from the first posting. Where is the 35' 8" supposed to go. (Yes, I'm a math nerd type ... but I speak English.)


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## Paul B (Mar 10, 2007)

Does your drawing show the 35'8 height at the corner or the middle of the wall?

Sorry wolf1728 I thought you were the OP.


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## wolf1728 (Jun 18, 2011)

Okay, I've redrawn it according to what I think it should be:









I'm thinking that all those numbers can't be exactly right.
Which 2 must be exact? (I'll do the exact calculation for the third).


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## Railman (Jan 19, 2008)

Wolf,
I'm glad you posted that. Something isn't right.
Here is a copy of an excell arc calc that I've had on my desktop for a while, but never quite got the hang of till now. I don't use Excell. I usually use an old dos based (similar to Lotus) in my daily spreadsheets. 
Anyway, Here's what I've got. It appears to check out given that the
ht at 1/2 of the 160 ft cord equals the 35.66 ft ht the OP stated. The rad. appears to be different than that originally stated though. It's 107.55ft or 107' - 6 5/8".

I suspect this is more in line with some of the earlier posts given with reference to cad program dimensions.
Hope this downloads, It's my 1st spreadsheet referal!:whistling

View attachment excell arc_calc MODIFIED RLMN.xls

edit: this spreadsheet is probably WRONG!
See later post below for updated values, & explanation.

I'd like to give credit, & thanks to Ron Zukerman for this great spreadsheet! It took me a while to figure where I originally found it. He's a woodworker, & has his own web site.
http://home.comcast.net/~ronzu2/mainpage.html
Joe


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## GettingBy (Aug 19, 2010)

Here's part of the calculation.

If
L is the chord length
h is the arc height above the chord in the center
and R is the radius of the circle that the arc is part of, 
then
R= {(L^2)/4 + h^2}/2h

So for a chord length of 6' and a maximum arc height above the chord of 1', the radius of the circle is
{(6^2)/4 + (1^2)}/2
={36/4 + 1}
={9+1}/2
=5
For this special case it's two 3:4:5 triangles inside a circle, with the 4' sides touching, so it's a triangle with sides equal to 5, 5 and 6. 
You can figure h just by looking at it.

For your case I get 
L = 160 x 12 = 1920" and 
h = 35' 8" = 428" so 
R = 1076.89" = 89.74' = 89' 8.89".




On an X-Y coordinate system, for this case, when 
at X = 0 [the height of the arc], Y should = 1 [the height of the arc]
and when X = +3 or -3 [the edge of the arc] Y should = 0 [the height of the end of the arc].

What happens when Y = 1/2?

For any values of X and Y using this system, X is given by 
x = sqrt([R^2] - {[y + R-h]^2})

so for the case above
= sqrt([5^2] - {[0.5 + 5-1]^2})
= sqrt(25 - {[0.5 + 4]^2}) 
= sqrt(25 - {[4.5]^2})
= sqrt(25 - {20.25})
= sqrt(4.75)
= 2.179449472'
So the X arc coordinate at Y = 1' is X = 2.179449472'
You can substitute any Y value in this formula and it will give you the X value.


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## wolf1728 (Jun 18, 2011)

As I said, I think the problem is with the 3 numbers that were stated in the beginning. 










Looking at the new diagram, we see that if the height (h)
equals 35.6666 feet then (what I call h2) must equal 
h2 = radius - h 
h2 = 181.9166666... - 35.66666...
h2 = 146.25
To check on this we use the Pythagorean theorem in which half the chord length squared (80^2 = 6,400) plus h2 squared (146.25^2 = 21,389.0625) must equal the radius squared (181.91666^2 =33,093.6736)
It does not since 27,789.0625 is less than 33,093.6736

Perhaps the problem lies in the chord length. Maybe this is what should have been calculated to begin with?


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