Thread: Quick/Dirty way to calculate volume...

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Quick/Dirty way to calculate volume...

Hey guys... so I had this idea, I have yet to test it but will when I get a chance.... on a way to calculate a roundabout general ballpark area for surboard volume.
This is probably used by people today but I have not read it anywhere...
so here are the steps...
1. run a string down the rail from nose stringer all the way down to tail stringer, taping the string if you need to on the rail, to keep it on the outtermost area of the rail. then do the same down the other rail.
2. attach the 4 ends of the string that meet at the tail(2) and the nose(2), either by taping or tieing them together.
3. get the thickness of the board at the tail, the nose area and the middle of the board.
then add them up and divide by 3 to get the average thickness of the board.
4. then remove the string from the board but still attached to itself where it meets at the nose and the tail and with 4 steaks or something to hold corners make the string into a square or rectangle shape(whichever you want) and use an online volume calculator like the one here to calcutate from h x w x thickness:
http://www.online-calculators.co.uk/...cubevolume.php
and you should be able to get a very close ball park of your boards volume...
would this work to get a close number you think?
I have a board that I know the volume of so I will test this when I get a chance.

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compare the board to a CI model, email them, they will supply volume for their model. or recreate the board in boardCAD and measure. L x W x H is volume of a cube. A surfboard is not a cube.

3. Swaylocks has a ton of discussion on this. The simplest suggestion I found without having to dunk your board in a pool or use some CAD tool, was the following (the results were only about 1 liter low for my Dumpster Diver):

"There is a great simple formula in the archives. I forget who posted it and I apologise for using it without their name attached (many thanks - I've found it very useful over the years, and pretty accurate). Here it is:

approx vol = 1/2Length x Width x thickness + (10 x every inch over 6') all measures in decimal inches

you get a big number (cubic inches)

divide by 61 to get liters

works best for shortboards"
Last edited by PhiloSurfer; Feb 7, 2013 at 08:46 PM. Reason: typo

4. Originally Posted by PhiloSurfer
Swaylocks has a ton of discussion on this. The simplest suggestion I found without having to dunk your board in a pool or use some CAD tool, was the following (the results were only about 1 liter low for my Dumpster Diver):

"There is a great simple formula in the archives. I forget who posted it and I apologise for using it without their name attached (many thanks - I've found it very useful over the years, and pretty accurate). Here it is:

approx vol = 1/2Length x Width x thickness + (10 x every inch over 6') all measures in decimal inches

you get a big number (cubic inches)

divide by 61 to get liters

works best for shortboards"
this is the best way i've found. as mentioned, it works best for shortboards. the closer to 6'0" the better. once you've divided by 61 (i actually think 60.02 is a touch more accurate, but whatever), if you divide the quotient by 28.32, it'll give you an approximate volume in cubic feet (cu.ft.), which is how mike & the guys at coil do their volumes.

ex: 6'4"x19.25"x2.5" CI Fort Knox = 31.7L

by the above formula, you'd get (38x19.25x2.5+40=1868.75/60.02=) 31.14L
31.14/28.32=1.1 cu.ft.

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Yeah I agree these are all proven ways... but wouldnt the steps I was thinking about would give a very close number on volume? I mean a 15 inch long string laid out as a square or a circle contains the same area, no?

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I'm not sure if the string method works. If you change the plan shape with the strings then you change the area, even if the length of the strings is constant. If the area is different, the volume is different. (Imagine the area of a circle created by the rim of a paper cup. Then squeeze it into an oval. Then squeeze the oval until the two sides touch. The area of this new shape is zero--or close to zero--even though the length of the rim hasn't changed.) I suppose there are ways around this problem....

7. just seems excessively complex/labor intensive when there are reasonably proven equations to get the same result. it doesn't have to be super precise...volume's just another # to be taken in conjunction w/ all the other dims.

8. Constant perimeter (the string) doesn't make for a constant area. If the string is 20 inches long you could make it into a a 5 in by 5 in square with an area of 25 square inches. You could also make the string into a 8 in by 2 in rectangel with an area of 16 square inches. The circle will result in the largest area. Just saying.

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Originally Posted by silas
Constant perimeter (the string) doesn't make for a constant area. If the string is 20 inches long you could make it into a a 5 in by 5 in square with an area of 25 square inches. You could also make the string into a 8 in by 2 in rectangel with an area of 16 square inches. The circle will result in the largest area. Just saying.
Its a 3dimensional process... wouldnt the thickness would be greater on the rectangle than the square?.... the only way I can know is if I try and see if the method matches one of my surfboard which I already know the volume of. Will give it a go when I can.
thanks for the thoughts on the matter too!
Last edited by MFitz73; Feb 8, 2013 at 01:16 PM.

10. You guys should talk to Mike Daniel's on the Coil Ride Report thread over at Swaylocks. He's got volume dialed in.