Calling All Mathemagicians!

[Crowbob]

Let’s say you have a shallow arc, the ends of which are 80-1/4” apart and the apex or deflection or height or whatever of the arc is 2-1/8”.

What is the total length of the line of the arc so described.

Extra credit if you show your work.

[masraum]

That's a really good question. Interesting.

Google says

To find the radius (R) of a circle using an arc, the most practical method involves measuring the chord length (W, the distance between the arc's endpoints) and the sagitta (H, the height at the midpoint of the arc). The formula is R=W/2+W^2/8*H

So 80.25/2 + 80.25^2 / 8*2.125 = 418.95"

[fanaudical]

https://en.wikipedia.org/wiki/Circular_segment

[Crowbob]

So the length of the arc segment is 80.3732”?

I’m calling it 80.5”.

Thank you!

[masraum]

Originally Posted by Crowbob

Let’s say you have a shallow arc, the ends of which are 80-1/4” apart and the apex or deflection or height or whatever of the arc is 2-1/8”.

What is the total length of the line of the arc so described.

Extra credit if you show your work.

Originally Posted by Crowbob

So the length of the arc segment is 80.3732”?

I’m calling it 80.5”.

Thank you!

In your first post you said 80 ¼" which would be 80.25 which is what I used for the calculations.

[Crowbob]

I used the number you came up with (418.95”) as the radius. Then I went on the interwebs and found a calculator and plugged in 418.95 for r (radius), and 80.25” for c (chord) and it told me the l (length) of s (segment) is 80.3732”.

I thought that didn’t look right. However, if you consider the circumference of a circle with a radius of 418.95” is 2632.34” or about 219.4 feet, it makes sense.

[masraum]

Originally Posted by Crowbob

I used the number you came up with (418.95”) as the radius. Then I went on the interwebs and found a calculator and plugged in 418.95 for r (radius), and 80.25” for c (chord) and it told me the l (length) of s (segment) is 80.3732”.

I thought that didn’t look right. However, if you consider the circumference of a circle with a radius of 418.95” is 2632.34” or about 219.4 feet, it makes sense.

Ah got it. Yeah, and that's good, because working back came up with basically the same measurement.

[Bill Douglas]

1.2 meters.

[masraum]

Originally Posted by Bill Douglas

1.2 meters.

Since his tiny portion of the circle is over 2m long...

[Bill Douglas]

I always say 1.2 meters if I haven't got a clue.

It used to be

Me: How long it it going to take?

Them: How long is a piece of string...

Me: 1.2 meters, now answer the question.

[zakthor]

Well... I thought about this and tried to come up with a cool solution using infinite series but I failed so here's a regular way and I can show my work.

The height from the middle of the line to the circle is H.
The distance between the two points (P0, P1) on the arc is D.

D/2 is the length to the middle of the span, I'll call that point D2.

The distance from that point in the middle of span up to the circle is H and I'll call that point H2.

The distance from H2 to the center of the circle is R - the radius of the circle.

Therefore the distance from D2 to the middle of the circle is... R - H.

So now we have a right triangle and two lengths:
P0 to D2 is length D/2.
P0 to middle of circle is R.
D2 to middle of cicle is R-H.

By pythagoras: R^2 = (D/2)^2+(R-H)^2

Expand to:

R^2 = D^2/4 + R^2 - 2RH + H^2

Cancel the R^2

0 = (D^2)/4 - 2RH + H^2

2RH = (D^2)/4 + H^2

R = (D^2)/8H + H/2

We have the circle's radius, now we need to know the angle between P0, P1 which is a trig identity:

Theta = 2 asin(D/2R)

Theta = 2 asin(D/((D^2)/8H + H/2))

Arc length is: R Theta

Plugging it all in you get:

S = ((D^2/4H) + H) arcsin(D / ((D^2 / 4H) + H))

S = 80.25^2/(4 x 2.125)+2.125 x arcsin(80.25/((80.25^2 / (4 x 2.125))+2.125

S = 80.45"

The arc is 0.2" longer than the distance of the span.

I'm still thinking about a different way to do this but probably its a dead end. Anyway there you go!

[red 928]

[Crowbob]

Good thing I didn’t ask how many kilograms are in a mile at 34°C during a lunar eclipse in Chatanooga simply because I’ve always wanted to know.

Hooboy, I love this place!

[Bill Douglas]

Originally Posted by Crowbob

Good thing I didn’t ask how many kilograms are in a mile at 34°C during a lunar eclipse in Chatanooga simply because I’ve always wanted to know.

Hooboy, I love this place!

Again, 1.2 meters

[Pazuzu]

Is this a hanging rope or chain or something? Because the calculations would be different if it is...

[Crowbob]

I needed to know if I had enough copper flashing for this ridge board: