45's

[Colossus281 10+ year member]

Is there a way to measure 45 displacement?

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[wenn_du_weinst 10+ year member]

yes, you measure it as a triangle.

 

[dbeez 5,000+ posts]

Kiddies this is where trig comes in handy take math seriously in school.

 

[Colossus281 10+ year member]

Wow what the hell is wrong with me?

 

[Kangaroux 5,000+ posts]

Wow what the hell is wrong with me?

Probably a lot. Brotip: if you are using sketchup, design the whole box on one sheet of wood using just lines to outline where everything will go. Put a line to mask off the end of the port, and then select the text tool and click the area where the actual chamber of the box is. Multiply whatever number you get by your box height and then divide by 1728. That is your net volume

 

[Colossus281 10+ year member]

Probably a lot. Brotip: if you are using sketchup, design the whole box on one sheet of wood using just lines to outline where everything will go. Put a line to mask off the end of the port, and then select the text tool and click the area where the actual chamber of the box is. Multiply whatever number you get by your box height and then divide by 1728. That is your net volume

Nope doing it old school. Pin and paper no design software.

 

[Buck 5,000+ posts]

(side ^2)/2 * height /1728

Pretend it's a square and cut the volume in half, it's easy. A side is how far the wood goes away from the corner.

 

[jeremyg1 10+ year member]

It is nothing more than an isosceles prism

If we take what Buck said above:

(side ^2)/2 * height /1728

and assign the following:

A = 10 inches

B = 5 inches

we get the following:

(((5^2)/2)*10)/1728

((25/2)*10)/1728

(12.5*10)/1728

125/1728

.072 cubic feet of displacement