Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

**fizzled****Member**- Registered: 2005-12-25
- Posts: 9

I have been working hard for hours to figure out how to calculate the volume of the geometrical body shown in the drawing below from just one parameter: the height of it's slant "s1". Finally I have figured out the formula.

V = ( r² b h l ) / s² - ( r³ b h c ) / 3 s³

I replaced "s1" in the drawing by r in the formula in order to better be able to write down the formula.

b, c, h, l and s have fixed values!

1. Question: Can somebody tell me if I did the formula right??

2. (most important) Question: How can s1 be expressed in terms of V?

in order to determine s1 in dependence of V one needs to read the formula this way:

" r = .... "

I don't know how to develop the formula so that I can read it this way!

May I please ask your help? I have no idea about polynoms and stuff like that.

----------------------------------------------------------------------------------------------------------------

CORRECTION ( SEE BELOW NO 6 ):

Volume was wrong. Correct volume might be: V = 2bhcr³/(3s³) + bh(l-c)r²/s²

Please check

*Last edited by fizzled (2005-12-27 11:04:23)*

Offline

**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,562

Is the back triangle slanted inward on the bottom, thus reducing the volume a little?

Is the front triangle side exactly vertical?

**igloo** **myrtilles** **fourmis**

Offline

**fizzled****Member**- Registered: 2005-12-25
- Posts: 9

John E. Franklin wrote:

Is the back triangle slanted inward on the bottom, thus reducing the volume a little?

Is the front triangle side exactly vertical?

Hi,

thank you for asking.

Yes, the back triangle (back border "wall" of the volume to be determined) is slanted inward on the bottom.

Yes, the front triangle side is exactly vertical.

Offline

**fizzled****Member**- Registered: 2005-12-25
- Posts: 9

Could somebody please help?

Checking the volume formula is not so important;

I would be happy if question no 2 could be answered, as this is something that I despair of!!!

Thanks in advance

Offline

**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,562

Hi fizzled, I figured out part of the problem, but not the whole thing.

If there was a vertical wall at point c (C-wall) from the back where the triangle slants and

hits the bottom, then the volume from the C-wall back to the slanted triangle

is exactly 2/3 rds the volume of going all the way back from the C-wall.

**igloo** **myrtilles** **fourmis**

Offline

**fizzled****Member**- Registered: 2005-12-25
- Posts: 9

John E. Franklin wrote:

Hi fizzled, I figured out part of the problem, but not the whole thing.

If there was a vertical wall at point c (C-wall) from the back where the triangle slants and

hits the bottom, then the volume from the C-wall back to the slanted triangle

is exactly 2/3 rds the volume of going all the way back from the C-wall.

Hi John,

Thank you for answering. I am afraid that does not help much to solve my problem. The volume to be determined is defined by s1 (=r). I understand that your idea could be useful, though, if the volume of the whole body (defined by s) should be determined.

In order to determine the volume defined by s1 (=r), my own thought was to subtract the volume of the pyramid in the back from the volume of the prism defined by the front triangle and l. But in the meantime I found a mistake in my formula. It is not correct. I erroneously subtracted the pyramid defined by c1 from the prism volume defined by s1 taking the wrong length l (instead of the contracted length). So I had to correct the whole thing. The correct volume should be:

V = 2bhcr³/(3s³) + bh(l-c)r²/s²

Could someone please check if this is correct?

But stilll my second question remains of how to transform this equation so that it can be read like this:

r = ...

The difficult thing is r³ and simultaneously r² in one equation.

That's too much for me.

fizzled

*Last edited by fizzled (2005-12-30 12:28:50)*

Offline

**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,562

I will use big L instead of little l, so you don't think it is a one.

Since we are told b=e, then the

Volume = he(L-c) + (2/3)hec

Then use pythagorean theorem perhaps to do

part 2, just guessing.

Sorry I'm doing the big volume with S, but I don't really

know what is the difference between the big one and

the small one.

Just multiply the big one's volume by (s1/s)^3 to get the small ones.

*Last edited by John E. Franklin (2005-12-27 09:55:52)*

**igloo** **myrtilles** **fourmis**

Offline

**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,562

Woops, I think that wouldn't work, because the length isn't changing proportionally.

**igloo** **myrtilles** **fourmis**

Offline

**fizzled****Member**- Registered: 2005-12-25
- Posts: 9

John E. Franklin wrote:

I will use big L instead of little l, so you don't think it is a one.

Since we are told b=e, then theVolume = he(L-c) + (2/3)hec

Then use pythagorean theorem perhaps to do

part 2, just guessing.Sorry I'm doing the big volume with S, but I don't really

know what is the difference between the big one and

the small one.Just multiply the big one's volume by (s1/s)^3 to get the small ones.

Thank you, John

I tried the two different formulas. Your formula works with determining the big volume and gives the same result as the one I figured out. But applying your factor (s1/s)^3 does not show the same results! So one of the formulas must be wrong?

Did I understand right? (s1/s)^3 = (s1/s)³ ??

fizzled

Offline

**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,562

I'm wrong about the one dimension ratio cubed because the length doesn't get shorter the same as everything else.

But if you could figure out how far the length is off, since it would be too short for the smaller similar shape of the

large one, then you could add on the prism slice to the front to make it the right length. But I don't know how yet.

I reread that, and it is unclear. If you reduce the big volume by the (s1/s)^3, then all the dimensions are right

except the length dimension, which is too short. But how far off is it??

As s1 gets smaller than s, c grows from zero upward, what is this equation?? Should be simple, but I am

not thinking of it. So when make the 3-d object reduced in size, length of object is reduced by (s1/s),

however, the object slides down the slant left and frontwards by c, so how is c related to s1?

*Last edited by John E. Franklin (2005-12-27 10:26:57)*

**igloo** **myrtilles** **fourmis**

Offline

**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,562

I have a hunch if c > L/2, then when you reduce the object, the bottoms length gets too long.

Just a guess.

If c = L, then there is no bottom, just a point and the reduction should work fine, maybe.

If c is much smaller than L, then... who knows, it's all proportions somehow, I gotta think more...

I think that was all wrong thinking.

When I said the reduced image slides down the slant to hit the bottom, that is not really

correct. If you reduce the image holding the bottom C point at a fixed location, then

everything reduces right, but the length is just too short by (s1/s), so this shouldn't

be too hard...

*Last edited by John E. Franklin (2005-12-27 10:47:11)*

**igloo** **myrtilles** **fourmis**

Offline

**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,562

Okay, tack on the front a slice of thickness (s - s1)/s

The reason I say this is because if s1=8 and s=9, then the new proportion is 8/9,

so then the part we need to add to the reduced length is 1/9th.

So (9-1)/8, hence (s-s1)/s

*Last edited by John E. Franklin (2005-12-27 10:53:21)*

**igloo** **myrtilles** **fourmis**

Offline

**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,562

So the volume of the front slice to correct for the reduction would be:

(L-c)((s-s1)/s)h1b1

**igloo** **myrtilles** **fourmis**

Offline

**fizzled****Member**- Registered: 2005-12-25
- Posts: 9

My ideas were:

h1 = h*s1/s

b1 = b*s1/s

c1 = c*s1/s

(theorem on intersecting lines)

*Last edited by fizzled (2005-12-27 11:21:33)*

Offline

**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,562

Sounds good. I think with result from post #13 and your #14, you might be almost there, though it might be

a little messy. And I hope this reduction idea is all right. I think it is.

**igloo** **myrtilles** **fourmis**

Offline

**fizzled****Member**- Registered: 2005-12-25
- Posts: 9

Can somebody change this formula: V = ar² + br³ into r = ... ???

Thank you for your help.

Offline

**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

Yes, I can, but it's ugly:

a=bx^3+cx^2

*Last edited by krassi_holmz (2005-12-29 11:19:45)*

IPBLE: Increasing Performance By Lowering Expectations.

Offline

**fizzled****Member**- Registered: 2005-12-25
- Posts: 9

krassi_holmz wrote:

Yes, I can, but it's ugly:

a=bx^3+cx^2

Hi krassi_holmz,

THANK YOU VERY, VERY MUCH for this transformation!!

I tried the formula immediately. But unfortunately it does not work for my a, b and c; excel does not calculate square roots from negative values!

My values are:

0,0448036323381 < a < 12,132000000000

b = 0,2135347303000

c = 1,7159763313610

applied to your result of a=bx³+cx² shown above.

I am helpless again.

Do you know if there is a solution to this problem??

Thanks again in advance!

*Last edited by fizzled (2005-12-30 12:33:54)*

Offline

**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

I'm sorry. Actually this runction has three roots.

1-real and 2-complex.

Here's it's roots:

*Last edited by krassi_holmz (2005-12-30 22:00:35)*

IPBLE: Increasing Performance By Lowering Expectations.

Offline

**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

Simplified:

*Last edited by krassi_holmz (2005-12-30 22:02:48)*

IPBLE: Increasing Performance By Lowering Expectations.

Offline

**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

I'll try something...

IPBLE: Increasing Performance By Lowering Expectations.

Offline

**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

Real solutions:

*Last edited by krassi_holmz (2005-12-30 22:10:16)*

IPBLE: Increasing Performance By Lowering Expectations.

Offline

**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

Simplifyed:

*Last edited by krassi_holmz (2005-12-30 22:12:49)*

IPBLE: Increasing Performance By Lowering Expectations.

Offline

**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

But we have conditions for a,b and c:

*Last edited by krassi_holmz (2005-12-30 22:39:30)*

IPBLE: Increasing Performance By Lowering Expectations.

Offline

**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

I was right from the begining!

If x=Root[a == b x^2 + c x^3,1](means the first root of f(x)) then Im[x]->0!

IPBLE: Increasing Performance By Lowering Expectations.

Offline