Math help - cubic feet
 

OK, I need some help from the rbp math whizzes. I'm trying to figure
out the cubic feet volume to a triangular enclosure in the shape of a
traditional A-frame tent (but a bit shorter).

Dimensions are 7' wide at the base, 7' from the base to the peak and 4
1/2" long.

How many cubic feet is that? I'm stumped.

In article ,
(Mike McCrea) wrote:

OK, I need some help from the rbp math whizzes. I'm trying to figure
out the cubic feet volume to a triangular enclosure in the shape of a
traditional A-frame tent (but a bit shorter).

Dimensions are 7' wide at the base, 7' from the base to the peak and 4
1/2" long.

How many cubic feet is that? I'm stumped.
Did you mean 4 1/2 inches or feet?

volume = length * (width * height / 2)

if feet then 4.5' * ( 7' * 7' / 2 ) = 110.25 cu ft
if inches then 0.375' ( 7' * 7' / 2 ) = 9.1875 cu ft

Dave Filpus wrote in message
if feet then 4.5' * ( 7' * 7' / 2 ) = 110.25 cu ft

Thanks Dave,

That was all in feet (7x7x4.5). I filled a bit less than half of the
available space in our new A-frame woodhshed last weekend with
firewood and wondered how close to a cord (128 cu ft) I came.

Look like the shed will hold 2 cords if filled completly.

On 8 Nov 2004 04:11:34 -0800, (Mike McCrea)
vaguely proposed a theory
.......and in reply I say!:

remove ns from my header address to reply via email

You've seen the result. The formula is 1/2 the base width, times the
_vertical_ height, (this is the area of a triangle)......

.....times the length. Imagine infinite triangles laid side by side to
make up the rectangular side.

OK, I need some help from the rbp math whizzes. I'm trying to figure
out the cubic feet volume to a triangular enclosure in the shape of a
traditional A-frame tent (but a bit shorter).

Dimensions are 7' wide at the base, 7' from the base to the peak and 4
1/2" long.

How many cubic feet is that? I'm stumped.

************************************************** ***
Dogs are better than people.

People are better than dogs for only one purpose. And
then it's only half of ofthe people. And _then_ most
of them are only ordinary anyway. And then they have a
headache.........

Since you are dealing with a triangle, the easy way is to drop a line
from the apex, cut it in half, and turn one half upside down and flop
it over the otherside. You now have a rectangle 3 1/2*7 or 24.5
square feet for the area of the end. Then multiply that by the length
in feet and you have the volume. if the length is 4.5 inches = 4.5/12
or .375 foot, the volume is 24.5x.375=9.1875 cu ft.
Or if you really meant 4.5 feet, 110.25 cu ft.
B=triangle base length * = times
H=height of triangle
L=length of object
V=1/2*B*H*L
Dan

On Mon, 08 Nov 2004 14:06:59 GMT, Dave Filpus
wrote:

In article ,
(Mike McCrea) wrote:

OK, I need some help from the rbp math whizzes. I'm trying to figure
out the cubic feet volume to a triangular enclosure in the shape of a
traditional A-frame tent (but a bit shorter).

Dimensions are 7' wide at the base, 7' from the base to the peak and 4
1/2" long.

How many cubic feet is that? I'm stumped.
Did you mean 4 1/2 inches or feet?

volume = length * (width * height / 2)

if feet then 4.5' * ( 7' * 7' / 2 ) = 110.25 cu ft
if inches then 0.375' ( 7' * 7' / 2 ) = 9.1875 cu ft

7x7/2*4.5=110.25
"Mike McCrea" wrote in message
om...
OK, I need some help from the rbp math whizzes. I'm trying to figure
out the cubic feet volume to a triangular enclosure in the shape of a
traditional A-frame tent (but a bit shorter).

Dimensions are 7' wide at the base, 7' from the base to the peak and 4
1/2" long.

How many cubic feet is that? I'm stumped.