# High School Geometry !!



## JackM (Jul 29, 2008)

This question doesn't seem to fit in any forum known to man, so I'll ask it here.

C
| 
A ______________|________________ B

I have a flat object 48 inches long. It's end points are at A and B. If point C, the mid-point of the flat surface is raised 4 inches creating a 48 inch arc of that surface, what will be the flat distance from A to B ? In other words, if I take a 48" piece of flexible material and make it into an arc with the mid-point (C) 4" high, what will be the distance between A and B?

The curved line A-C-B is 48". What is the distance A to B along the flat surface that originally was 48"?

I'm sure any high school Sophomore can do it. I'm lost.

JackM


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## wchasr (Jan 2, 2008)

Approx 47 1/8". I cheated and drew an arc in AutoCAD and adjusted it until it fit the arc length of 48 inches. The radius becomes about 71 3/16". I took a line 48 inches long and offset it 4 inches. Drew an arc from endpoint to mid point and then to endpoit and stretched it until the arc length was about 48 inches. It's not exact but close. 

Chas


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## JackM (Jul 29, 2008)

Wha' ?? 

Doesn't seem like much, but I'll accept your answer. Thanks. 
JackM


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## Del Tapparo (Jan 4, 2008)

There is more than one solution to this problem, because the unknown variable is the radius of the circle.

Let C = chord length (trying to solve for this)

Let H = the deflection from the chord to the arc = 4
Let R=the radius of the circle
Let A=angle of the chord in radians
Let S = the arc length = 48 

pi = 3.14159 


C = sqr root of (4H(2R-H)
S=RA

so C = sqr root of (4H(2S/A)-H)

For A of pi (180 degs)
C=sqr root(16(96/pi)-4) = 20.6

For A of pi/2 (90 degs) C = 31.2

For A of pi/4 (45 degs) C = 44.2 


The smaller the angle, the closer C gets to S

This is driving me nuts too. Although this is what the equations say, it still doesn't sound right. Anyone else? 


I think it makes sense if you draw a couple of examples with a large and small chord angle.


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## Gary Armitstead (Jan 2, 2008)

I tried to construct this in Mastercam, but to no avail. My wife taught this stuff until about two months ago, when she retired. I'll ask her when she gets home.


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## wchasr (Jan 2, 2008)

Jack I used the Drafting program on my computer at work to do the calculations. No it's not exact but it's close. It's not even proper geometry and my Grandmothers are spinning in their respective graves (both having been Math teachers). I drew a line 48 inches long. I offsetted it 4 inches to give the 4 inch offset knowing that the centerline of the arc would be thru the center of the lines. I then drew an arc from the end point of the lower line on the left (your point A) to the midpoint on the upper line (your point C) and back to the endpoint on the right of the lower line (your point B) now this gives us an arc of an un-known radius (although the computer tells me that it is 74 inches) I then draw a line from "A" to the center of the arc and back to "B". I also draw a center line from "c" to the center of the arc. I ask the computer what the arc length is at this point and it is 48.884 (resolution defaults to 3 decimal points) so I offset the line from "A" to center & from "B" to center by about 7/16 or .4375 I then stretch the arc to the intersections of the offset lines with our lower line. This gives me the results I reported to you. It's not truly accurate but close. Hopefully close enough? 

Chas


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## toddalin (Jan 4, 2008)

Yes, ~47-1/8"

Keep it simple and use 90* triangles and it comes to 47.3", but the curve takes a bit more length around the arc than two right triangles across the hypotenuse, so ~47-1/8" it is.


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## Michael Glavin (Jan 2, 2009)

Jack, 

I get 47.10" length from a 48" line with a 4" arc offset. 

Michael


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## SteveC (Jan 2, 2008)

It may not be an elegant method, but it's how I backed into the answer.


















I went here Circle Sector, Segment, Chord and Arc Calculator[/b] (scroll down to calculator).

Selected the known values as Segment Height & Apothem and kept increasing the Apothem value until I got to an arc length of 48.


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## Del Tapparo (Jan 4, 2008)

Great find Steve. I kept running into the intersecting chords theorem, but the trick is; the circle's diameter is also a chord. They show the formulas here if anyone is interested. Chord Formulas


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## xo18thfa (Jan 2, 2008)

Wow Steve!! Thanks for that link. It's bookmarked already. I had been using this website for segments of a circle, but the arithmetic makes my hair hurt and teeth itch.

Dr. Math


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## JackM (Jul 29, 2008)

I feel like I'm next at Pin the Tail on the Donkey: blindfolded and spun around three times. 

I think the answer I need is in there somewhere. Basically, "not very much". This weekend I expect to start construction of my train shed, patterned after the St. Louis Union Station train shed. Picture a raised flower bed kind of "box", filled with soil, gravel and ballast, with tracks about even with the top of the cedar edges. The curved top is a sheet of Lexan, with additional lengths of Lexan along the sides and ends to allow complete visibility of the rolling stock. This "yard" will allow a modest amount of switching in and outside of it, while providing a high level of security for storage of my all of my equipment. 

If I can pull this off, it should be pretty good lookin'. I'll take pictures..............unless it turns out to be a real disaster. I hope I can have it finished before the snow flies. 

JackM


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## Madman (Jan 5, 2008)

Now that all the rocket scientists have had their way with you, I'll tell you to take a piece of thin wood measuring 48"in length. Lay it flat on a table. Stabilize the ends, and push the center over 4". Now measure the distance from one end of the stick to the other on a straight line, or chord. Amazing! Better known as Chords for Dummies.


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## Semper Vaporo (Jan 2, 2008)

Posted By JackM on 15 Sep 2010 08:06 PM 
I feel like I'm next at Pin the Tail on the Donkey: blindfolded and spun around three times. 

I think the answer I need is in there somewhere. Basically, "not very much". This weekend I expect to start construction of my train shed, patterned after the St. Louis Union Station train shed. Picture a raised flower bed kind of "box", filled with soil, gravel and ballast, with tracks about even with the top of the cedar edges. The curved top is a sheet of Lexan, with additional lengths of Lexan along the sides and ends to allow complete visibility of the rolling stock. This "yard" will allow a modest amount of switching in and outside of it, while providing a high level of security for storage of my all of my equipment. 

If I can pull this off, it should be pretty good lookin'. I'll take pictures..............unless it turns out to be a real disaster. I hope I can have it finished before the snow flies. 

JackM 

I know how you feel!







I had a similar problem in building my elevated layout and went in search on the web for an answer and found that there is no way to get the answer given only the two values you know (length of arc and height of arc). You need one more numerical value to get a formula to calculate the chord. So, these folk here have found ways to guess at a third value such as to finally give you the answer you seek. The iterative method used by SteveC works fine.

Although I like the right-triangle method Toddalin used to estimate it, i.e.: assume it to be a "tent" with straight sides, the curve being approximated by the hypotenuse of the right triangle of 1/2 of the Lexan sheet:

Approximate Chord = 2 * sqrt( (48"/2)^2 - (4"^2) ) = 47.3286 but since it is really a "curve" that will shorten the value some more.

The estimate will be very close considering that I doubt of the Lexan will produce a true circular curve, but will produce a catenary arc, parabolic arc or some other non-circular arc.

My problem was "cured" because as I was searching for the formula so I could write a program to calculate values to fit a curved track on straight boards, I ended up forcing another dimension to a known value which then was enough for a simple forumla to calculate the chord. My program still had an iterative sequence to get one of the answers I needed and I always felt kind of stupid for having to write it like that, until I finally found a web site that explicitly said it could not be done in a simple forumla. Then I felt pretty smug to have done it the way I did!

Although the program worked well, my construction capability failed to meet the precision necessary to have needed the program to calcuate the answers at all!


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## SteveC (Jan 2, 2008)

Posted By JackM on 15 Sep 2010 08:06 PM 
I feel like I'm next at Pin the Tail on the Donkey: blindfolded and spun around three times. Tom

I don't know why, if you look at my previous reply.
[*] If you've got an arc (curved line AB on the outside of the green area of diagram) which is 48" in length.

and

[*] A segment height (straight vertical line ED in the green area of diagram) of 4"

then have to have

[*] A circle with the raidus (line OB or OA in the diagram) with the length of 71.322"

That will result in

[*] A chord (straight horizontal line AB between the red & green area) with a length of 47.099" 
[/list] Which pretty much comfirms the answer that Michael Glavin gave you in his reply i.e. 47.1".


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## wolf (Sep 30, 2010)

Hi Folks,

Every now and then, I look to see who is linking to my website www.1728.com and almost always it is interesting.

I believe there are 21 combinations of circle parts (chord and arc, radius and chord) and I thought I didn't have to include _all_ 21 combinations. 

Well it seems I'll have to include a calculation for segment height and arc length. 

Incidentally, I calculated 47.0993 for the chord length so you folks got the correct answer.

By the way, you couldn't find the formula to solve for a circle when you know the segment height and arc length because there is no formula. (Just as there is no formula when you know chord and arc length.) How is it solved? Either by trial and error or a more sophisticated procedure such as Newton's Method.


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## SteveC (Jan 2, 2008)

Hey Wolf

Thanks for dropping by and thanks for the calculator on the 1728 web site.


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