Length of arch

[itemphoto@yahoo.com]

Ok all you math experts, heres my problem.
I have an arch 68inches long, 5 inches high. I need to find the length as if the arch were a straight line. I use Turbocad 8pro and insignia. I am thinking this can be done in tc, but not sure how. And HAPPY Holidays to all!!!!

Kevin itemphoto@yahoo.com (mailto:itemphoto@yahoo.com)

[mikejohn]

Kevin
If the 68 inches is the straight line distance between the two end points of the arc, then the arc length is 68.92 inches.
.......Mike

[mikejohn]

Kevin
On the other hand, if the length of the arc is 68 inches, the straight line distance between the points is 67.07 inches.
......Mike

[mrdovey]

Kevin...

If the arch is circular (not elliptical) you can determine the radius and included angle (see this page (http://www.iedu.com/DeSoto/CNC/) for how-to calculate) and then determine the length of the curved line as:
length = radius x angle (in radians)

HTH

Morris

[Paul A (Unregistered Guest)]

Hi Kevin

Artcam shows the arc length as 68.9761....Try finding that on your tape measure !!

Regards

Paul

[bill.young]

Kevin,

Don't know about TC8, but in TC4 you select the arc, pick SELECTION INFO from the VIEW menu, and one of the fields in that box that appears is "Arc Length".

Bill

[kevinr (Unregistered Guest)]

Thanks to all. As always a great bunch of people here. Morris and Bill I will look over your answers, Mike thanks for the answer, how did you do it, Paul how did you figure this in artcam?
Thanks guys!!
Kevin

[mikejohn]

Kevin
I was an Aerial Survey Navigator (map maker) in my previous life
I used geometry, similar to the link Morris pointed to.
I've just checked my calculations, having seen Pauls answer, and see I typed a 2 instead of a 7, sorry

Calculated to 4 decimal places does give 68.9761.
This also checks out in AutoCad 2000.
....Mike

[mikejohn]

A little Chritmas puzzle for all.
If you had a length of rope forty two million, two hundred and forty thousand yards long (42,240,000 yards) it would exactly stretch around the equator, touching the ground.
If a keen ShopBotter went around the equator, putting in fence posts every 10 yards (a lot of fence posts) 4 feet six inches high, how much more rope would you need to go along the tops of the post?
10 yards
10 miles
100 miles.

.......Mike

[Paul A (Unregistered Guest)]

Kevin

If you right click on the arc and then select
properties it will show the perimeter length

Paul

[fleinbach]

10 YDs

[mrdovey]

If the rope were stretched tight, it looks a lot like 10 yards.

I figured (42,240,000/PI + 3) * PI, assuming that the 4.5' measurement was the length of fencepost left above ground.

Is there a catch to this?

Morris

[bleeth]

Morris: If the rope were strung at the top of the post then it would be plus 9 to get new D. The trouble is the Earth is not smooth and by stretching your rope on post tops you would be taking out a lot of kinks. If the original rope length were a true length rather than extrapolated from an average diameter then there is no equation for the refiguring. Picture two poles 10 yards apart in a straight line from tip to tip with an 8 foot wide 100' deep chasm between them.
So the answer is-Not enough info given.
Dave

[mrdovey]

Dave...

I'm willing to agree with everything except your use of the number 9 - because the fenceposts were spec'd out at 1.5 yards (4.5 feet).

Your chasm argument is probably a good one; but I've never been accused of not being a good "straight man". (-:

...Morris

[beacon14]

OK, if you increase the diameter by 3 yards, you increase the circumference by 3*pi, or 9.425... yards.

However, as we ShopBotters are a picky bunch, you have to take into account that by raising the rope off the surface of the Earth, it will no longer describe a circle but rather a series of straight lines, each 10 yards long (actually a bit more than 10 yards, as the fence posts will be ever so slightly farther apart at the tops than at the surface), so you may have a few inches of rope left over.

I remember being amazed when I first heard this problem, stated a little differently: If you wrap a steel rail tight around the Earth, then cut the rail and add 6 inches, how far above the earth's surface will the resulting rail float?

[ron brown]

Open NURBS arc
start = (-34, 0, 0) end = (34, 0, 0)
degree = 2
control points: rational, count = 3
knots: uniform (delta = 68.9762), domain = 0 to 68.9762
arc center = (-3.55271e-015, -113.1, 0) radius = 118.1 angle = 33.4635 degre

From Rhino "object properties/information"

[mikejohn]

Actually, it was because I was amazed when I first found this out, I asked this 'fun' question.
It is, of course, only theoretical.
This planet is anything but regular.
Mapmakers even use different centres for the earth. It wasn't until NASA launched the GPS satellite system we had a uniform shape to work too.
So Dave R. The amount of information to give a true answer would be astronomical.
Try asking a mapmaker what a straight line is!
Another small Christmas puzzle.
One new years day I walked from Boston to New York in under an hour. How was this possible?

......Mike

[Paul A (Unregistered Guest)]

Hi Mike

The bus was full so you walked up and down the isle....you could say it only took 1/2 hour as you were walking back to Boston 50% of the time


Paul

[mikejohn]

Paul, good thinking, but no.
My feet were touching mother earth every step.
........Mike

[normand]

probably a different new york Boston somewere in uk

[mikejohn]

In fact the original Boston!


4247

4248

.....Mike

[Paul A (Unregistered Guest)]

Mike

I went to a tea party in Boston once and I was back in time for dinner ??

Paul

[bleeth]

Mike-Last time I was there you couldn't have done it in under an hour-It was way too cold!!
Happy Christmas to all.

Dave

[mrdovey]

Kevin started me thinking about cutting arches, coves, and rounds. I decided that as I worked my way through this kind of problem, I'd add what I came up with to an online resource "gallery".

There isn't much there yet; but there's an index at http://www.iedu.com/DeSoto/CNC/index.html

...Morris

[itemphoto@yahoo.com]

Morris thanks for the index. It will surely help people like me!

Kevin Reid