Ot- How to calculate the area of odd shapes?

Trboatworks · Apr 5, 2017

Hello all-

I need to get a area of the shape below so I can calculate the displacement of a length of this:

What is the quick and dirty method here?
Can I just get the circumference and pretend this is a cylinder?

Thanks

Larry Dickman · Apr 5, 2017

if your trying to calculate the displacement, why not fill it with water.

crossthread · Apr 5, 2017

If it will not hold water then submerge it in a tub of water filled completely to the top and measure the overflow. You can set the tub in a shallow pan and measure the amount of overflow. It will equal the displacement.

powerglider · Apr 5, 2017

if you have the time, trace it on a piece of paper, take a picture, trace it by importing it in a 2d cad program. Scale it by using measurements taken from the object. Have the program calculate the area or volume by extruding it.

I've used Corel Draw and a feature they call "trace", then I would export it to cad program.

Madis Reivik · Apr 5, 2017

If the wall thickness is always same, then weight, calculate volume and divide by thickness and multiply by 2 to get total surface area.

Marty Feldman · Apr 5, 2017

powerglider said:
... trace it on a piece of paper, take a picture, trace it by importing it in a 2d cad program. Scale it by using measurements taken from the object. Have the program calculate the area or volume by extruding it.

Getting the area is what your polar planimeter is for.

I am curious as to what you are after, the displacement of the metal in the mast? The mast treated as if it were a solid?

-Marty-

Trboatworks · Apr 5, 2017

Marty- I need to weld up some pontoons out of this scrap mast section and was just trying to get an approximation of how much length I need for my buoyancy target.

Bobw · Apr 5, 2017

Unless the camera is playing tricks, I think you could approximate it as 2 trapezoids.
Or a rectangle and 4 triangles... The 4 triangles making up 2 rectangles. It would
be a little on the low side, but should be close.

D Dubeau · Apr 5, 2017

Me personally I would make a couple rough sketch's in cad, extrude it, and do a volume calculation. The "old" way would be to break it down into easy to calculate geometric shapes like the 2 partial circles on the side, and the rectangle in the middle, then math it out. My math skills have really degraded because cad makes it so easy for stuff like this, to the point that I don't even bother trying to solve things like this anymore longhand. It becomes an effort in frustration, because I don't remember equations (always had trouble anyway).

The other quick and dirty way would be to just fill it with water as someone said. tape plastic bags over any openings and fill it up.

rons · Apr 5, 2017

Trboatworks said:
Can I just get the circumference and pretend this is a cylinder?

No, you know what an ellipse.

Cross Sectional Area
A = 3.14159 * a * b

b = small_length / 2
a = large_length/2.
Then subtract a small factor for the clipped off ends. Approximate the clipped off ends as a half circle or a half ellipse.

Volume = A * Length_of_channel.

robert123 · Apr 5, 2017

If a standard extrusion, you may be able to get the lb/ft from the vendor or manufacturer. Divide by 12 to covert ft to inches and then divide by the density in lb/in3.

Otherwise, just weigh a length of it and divide by the length in inches. Divide this by the density. I am assuming you know what alloy it is.

Mtndew · Apr 5, 2017

If you have a cad/cam program, you could try drawing up an approximate part and analyzing the properties of it.

Scottl · Apr 5, 2017

When I was in school we were taught to calculate cutting force and center of pressure for cutting dies by breaking the shapes down into standard shapes stuck together.

A modern method would probably be to draw a rectangle representing the center portion and then attach 2 3-point arcs.

Real quick and dirty would be to attach temporary flat end caps to a length of it and submerge in a tub of water of known dimensions and measure actual displacement for a "Eureka" moment.

Trboatworks · Apr 5, 2017

rons said:
No, you know what an ellipse.

Cross Sectional Area
A = 3.14159 * a * b

b = small_length / 2
a = large_length/2.
Then subtract a small factor for the clipped off ends. Approximate it as a half circle or half ellipse. The

Volume = A * Length_of_channel.

Ok - the reason I thought that was figured in my minds eye if one were to run this section over a mandrel and deform to cylinder without changing Wall thickness you would have a size defined by how many inches of material runs around that mast section.

Am I really off in la la land on that?

I think I will line with a plastic bag and pour in some water to get this I just thought my circumference ideal was in the ballpark...

SteveF · Apr 5, 2017

Scottl said:
.....................
Real quick and dirty would be to attach temporary flat end caps to a length of it and submerge in a tub of water of known dimensions and measure actual displacement for a "Eureka" moment.

Or just put "temporary flat end caps" made of duct tape over the ends, put it in the water and use a bathroom scale (plus heavier known weights if needed) to see how much force it takes to submerge it and go straight to the final answer needed. Adjustment for weight of the mast already included!

Steve

dgfoster · Apr 5, 2017

Use the plastic bag.

the circumference will get you into trouble: reduced to simple case, if the original form were a 6x6" square section, its area is 36 and circumference is 24. Now change the cross section to 12" long by .001 wide. The cross section is .012 but the circumference is 24.002.

Denis

I always try to take cases to extremes to test hypotheses. Beats my slow calculation ability every time...

dsergison · Apr 5, 2017

Trboatworks said:
Am I really off in la la land on that?

It's a slippery slope to lala land. Imagine if you squashed it completely flat. it would have the same perimeter but no volume.

It's true your shape isn't squashed very flat, but how accurate did you want to be?

it's darn near a rectangle. better to approximate the width as sort of the average width.

Red James · Apr 5, 2017

For the accuracy you need, just wrap a tape around the outside and call it a a circle diameter. Divide by 2 to get radius, subtract rough thickness (1/2"?) to get inside circle radius, then pi*R*R for area. X length to get volume.

CalG · Apr 5, 2017

The area of a section can be obtained by tracing the shape onto graph paper and then "counting squares". The lions share of the squares will fall completely inside the shape you have. Estimating the "edges" isn't difficult. Not perfect, but pretty close, and doesn't require a computer or imaging devices.

Scottl · Apr 5, 2017

SteveF said:
Or just put "temporary flat end caps" made of duct tape over the ends, put it in the water and use a bathroom scale (plus heavier known weights if needed) to see how much force it takes to submerge it and go straight to the final answer needed. Adjustment for weight of the mast already included!

Steve

By "attach" I meant something stuck on with duct tape. IMO the most accurate method of determining displacement is to seal the ends and submerge it, trying to avoid submerging hands which will add to displacement somewhat. If the dimensions of a container are known and adding the piece raises the water level a measurable amount above the previously marked level the amount of water displaced can be calculated and converted to a displacement weight.

Ot- How to calculate the area of odd shapes?

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