Math Help Needed

[aramis]

Formula for a section of a cone, given
r1 r2 and h, where r1 is radius base and r2 is radius at h

(Reason why? I'm trying to work out the volume of the Serpent as shown in BOJTAS.)
(I think it's got ISSUES... 1.5m squares put it at 215T and 1.0m at 64T, not counting wings.)

 

[coliver988]

not sure if this helps, but you could start here: http://en.wikipedia.org/wiki/Cone_(geometry)

and pretty much any web search will yield pages that may have the info.

I'd try to help more but I'm not exactly sure what you are looking for (it's late for me, and I've a cranky son upstairs wanting me up there NOW).

 

[aramis]

No, it didn't... since I remember the formula for a right circular cone.

V=1/3(πr²)h​

Given a Right Circular cone of
unknown height X,
diameter A at base
diameter B at height H
find V without solving for X first

I know the long way, lets see if I remember how correctly:
solve for X
X=H*(A/(A-B))
and thus
V=1/3( (πA²X) -(πB²(X-H)) )
V=1/3π( (A²X) -(B²(X-H)) )
subbing in X
V=1/3π( (A²(H*(A/(A-B)))) -(B²((H*(A/(A-B)))-H)) )
And here's where my brain cramps...

 

[robject]

Even just counting squares alone, it seems to be 150 tons, so with the wings added I'd not be surprised if it were 200+ tons.

Can we assume the deck height is 2.5m instead of 3m? That gives us 83% volume.

I guess it's a cylinder though, not a slab, so that'll cut back on the volume a bit... hmph.

Slab, width w, height also w: length x w x w
Cylinder, diameter=width w: length x base; base = pi x w/2 x w/2, so: pi/4 x length x w x w = 78% slab volume.

So 83% of 78% of 150 is 97 tons, for the fuse.

Wings are free, apparently?

Ah, I see. The front is a cone. And the back tapers slightly. Maybe that'll buy us a few tons. Will it be enough to account for the wings?

I'm willing to say, "close enough", especially if we're within 10% of tolerance.

 

[Lycanorukke]

Hmm, it would be easier with images but I'll try and nut it out...

R1 - cone base radius
R2 - Cone radius at height H
H - Height given
X - total height of cone

Step 1:Find slope angle

Squash the cone into a triangle, and split it down the middle (2 right angle triangles)
Draw a horizontal line through the cone at H
Where the horizontal line intrersects the slope, drop a vertical line to the base
This forms a smaller triangle

The small triangle has a base of R1 - R2 (lets call it R3)
The height of this triangle is H

The angle of the base is Tan-1(H/R3)
You now have the slope angle of the cone (theta).

Step 2: Find the cone height

On the big cone triangle
Base is R1
Angle is theta
Height is X
So X/R1 = tan(theta)
X=R1 Tan(theta)
You now have the height of the cone and can work from there...

Hmm...as I said drawings are easier to describe

On edit:
Ah bugger, I just saw you didn't want to solve for X, and then work by subtracting volumes.
Crap. Sorry

 

[Tekrat04]

No, it didn't... since I remember the formula for a right circular cone.

V=1/3(πr²)h​

Given a Right Circular cone of
unknown height X,
diameter A at base
diameter B at height H
find V without solving for X first

I know the long way, lets see if I remember how correctly:
solve for X
X=H*(A/(A-B))
and thus
V=1/3( (πA²X) -(πB²(X-H)) )
V=1/3π( (A²X) -(B²(X-H)) )
subbing in X
V=1/3π( (A²(H*(A/(A-B)))) -(B²((H*(A/(A-B)))-H)) )
And here's where my brain cramps...

Your math is good. I just made a quick excel 2007 spreadsheet. PM me with your email if you want a copy.

 

[Starviking]

A Tangent

BOJTAS? Can you spell out the 'BO' part of that acronym?

 

[Lycanorukke]

BOJTAS? Can you spell out the 'BO' part of that acronym?

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