Little mathematics problem

[XabbaRus]

OK guys I want to model in 3DS MAx an accurate Roman scutum (shield)

I have used a size from wikipedia that says its width was .66m across and the cirumference around the curve was 0.86m. I am trying to work out what the original width was before being turned into a partial cylinder. Maths isn't my forte. Thanks for the help.

[antikristuseke]

the math for this is relatively simple, but i cant remember the formulas to be used. allso its time to sleep before work

[Platapus]

Lemme get this straight (pun intended)

You have a curved piece of metal. The distance between the two opposite edges as measured across the void of the curve is 0.66 meters

I am a little confused by the use of the term "cirumference around the curve was 0.86m." Do you mean the distance between the same two opposite edges as measured along the curved surface is 0.86 Meters?

You want to know if you flattened this curved piece of metal out flat, what would be the distance between opposite edges now measured along the flat surface?

Would this not be 0.86 Meters?

To make this shield one would take a piece of metal 0.86 meters wide and bend it with a constant radius until the arch formed would measure 0.66 meters across the void between opposite edges.

Or does "cirumference around the curve" mean the circumference of the virtual circle if the material of the shield were extended to make a complete cylinder of the same constant radius?

That does not seem right as the diameter of such a cylinder would be only 0.27 Meters in diameter which is pretty small for a shield.

I guess I am stubbing my eye on "cirumference around the curve".

[Letum]

I'm sure someone will come up with a mathematical answer, but personally I would just
use two guide splines at the correct length and change the width and bend of my
shape until it was close to the guide splines.

[XabbaRus]

You are right if i was making it for real I'd just need a sheet of metal .86 metres across and bend it until the distance across the void was .66 metres.

However in max if you bend something you end up changing its overall width.

[Prof]

I think the answer is that the original cylinder radius would have been 0.35m. The shield would cover 140 degrees out of the total 360 degree circumference.

To check:

(140/360)*2*3.142*0.35 = 0.855m (the length of the face of the shield)

2*0.35*sin(140/2) = 0.658m (the width of the shield)

Are those numbers accurate enough? I did the calculations numerically to make it quicker, but I could work out an analytical solution if you need more precision.

[XabbaRus]

Prof you have lived up to your name.

That is great cheers.

[Letum]

Originally Posted by [LEFT

XabbaRus[/left]]You are right if i was making it for real I'd just need a sheet of metal .86 metres across and bend it until the distance across the void was .66 metres.

However in max if you bend something you end up changing its overall width.

No, No, you misunderstand me!

I would make two splines .66m across and 0.86m. I would then change the width and
bend amount of the shield object until it matched the dimensions of the splines.
The final width of the sheild would be >0.86m when un-curved.

[bradclark1]

Originally Posted by antikristuseke

the math for this is relatively simple, but i cant remember the formulas to be used. allso its time to sleep before work

:hmm:

[antikristuseke]

Originally Posted by Platapus

Lemme get this straight (pun intended)

You have a curved piece of metal. The distance between the two opposite edges as measured across the void of the curve is 0.66 meters

I am a little confused by the use of the term "cirumference around the curve was 0.86m." Do you mean the distance between the same two opposite edges as measured along the curved surface is 0.86 Meters?

You want to know if you flattened this curved piece of metal out flat, what would be the distance between opposite edges now measured along the flat surface?

Would this not be 0.86 Meters?

To make this shield one would take a piece of metal 0.86 meters wide and bend it with a constant radius until the arch formed would measure 0.66 meters across the void between opposite edges.

Or does "cirumference around the curve" mean the circumference of the virtual circle if the material of the shield were extended to make a complete cylinder of the same constant radius?

That does not seem right as the diameter of such a cylinder would be only 0.27 Meters in diameter which is pretty small for a shield.

I guess I am stubbing my eye on "cirumference around the curve".

I just reached the same conclusion now that I thought about it.