Nubie ? Need help calc. vol. of room which is the sum of 2 diff. size triangles. (1 Viewer)

[John Pine]

I’ve forgotten my junior high geometry. The floor of my room is 23’ x 18’. But the ceiling is vaulted with a 14’ peak in the middle. The ceiling then drops vertically 3’ for four horizontally mounted skylights and then angles to other side of the room. So, basically if I can calculate the volume of these two triangles I should be able to just add the sum. Right? What’s the formula?

 

[ScottHH]

I don't really follow the explanation of your room in your post, so I can only give you the formulas I think you need, otherwise I’d do the math too. As you noted, you need to "split" your room into manageable segments to calculate each part's volume, and sum the parts to get the total.

The base of the room is a cube. It is 23x18x(height to bottom of vaulted ceiling). Multiply these three numbers.

A triangle is a flat shape. The area of a triangle is ½*base*height. Make sure you are using the two smaller sides as your base and your height (not the longest side aka the hypotenuse). Multiply the area of this flat triangle by the length, and you’ll have the volume of the area. FWIW, this website
tells me the shape is actually called a triangular prism.

If the vaulted ceiling was only one piece, this would be the calculation:
length depth height volume
Room(cube) 23 18 8 3312

length base height volume
Triangle1 23 9 6 621
Triangle2 23 9 6 621

Total 4554

 

[John Pine]

ScottHH: Thanks for the feedback. My description is probably not very accurate. You're right, I should calculate the volume of the two "cubes", which make up the lower portion of the room. The 14' peak is actually in the middle of the 23' length. So the 23' length should be cut in half. Then I should measure from the floor to where the angle of the ceiling starts. I think this is generally 8' or 9', but I'll check that tonight. I'll then be able to get the volume of the two cubes, which make up the lower portion of the room. LxWxH, Right? Then, add the volume of the triangles prisms to the two cubes and this should be correct. Right?

 

[Nathan Stohler]

The volume is width * length * height. The height varies, but since the peak is in the middle, we know the average height is (14 + 11)/2 = 12.5 ft.

V = 23 * 18 * 12.5 = 5175 ft3

You could also calculate the volume as two triangular prisms, but since the peak is in the middle, it doesn't even matter how the peak divides the room:

V = 2 * [ 11.5 * 18 * 12.5 ] = 5175 ft3
or
V = 2 * [ 23 * 9 * 12.5 ] = 5175 ft3

Note: I didn't account for the skylights

 

[John Pine]

Ok...I'm following Nathan. So the avg of the two triangles heights, should give a close enough estimate. Correct?

 

[Nathan Stohler]

Actually, it's more than just an estimate; it's the right answer. Keep in mind that I didn't subtract any volume for your skylights that drop down since I don't know their dimensions...

Also, I think I misunderstood your original post. Instead of 11, you should use 8 or 9 or whatever the height of your ceiling is at the edges. I thought you were saying the vault sloped down a total of 3 feet.

So, it should've been (assuming 9' ceilings):

V = 23 * 18 * (14 + 9)/2 = 4761 ft3

 

[ScottHH]

I'm not sure why you say there are 2 cubes in the lower portion of the room. But yes, volume of a cube is LxWxH. And I agree, if you add the volume of the cubes and the triangle prisms you'll get the total volume of the room.

I think Nathan posted his answer before your second post. But his shortcut of using the average height of the triangular portion of the room makes the calculation nice and easy. If the average height of the ceiling is (14+8)/2 = 11 ft, his calculation will not-surprisingly total to 4554 ft3.

It really comes down to how irregular your vaulted ceiling is, and how precise you want the measurement. Just continue to breakdown the wall into rectangles and triangles. Calculate the area of each of the shapes, and multiply the area of each of these by the other dimension of the room (18'). Add them all up.

 

[John Pine]

Got it! Actually the "skylights" are just four small 2'x 1.5' windows mounted on that vertical plane. So they should have no impact on the volume. Thx Nate and Scott!