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

You are not logged in.

#1 Re: Help Me ! » Extremely complicated change of a volume formula » 2005-12-30 12:18:30

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!

#2 Re: Help Me ! » Extremely complicated change of a volume formula » 2005-12-29 11:02:29

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

Thank you for your help.

#3 Re: Help Me ! » Extremely complicated change of a volume formula » 2005-12-27 10:57:50

My ideas were:

h1 = h*s1/s
b1 = b*s1/s
c1 = c*s1/s

(theorem on intersecting lines)

#4 Re: Help Me ! » Extremely complicated change of a volume formula » 2005-12-27 10:09:04

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 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.

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

#5 Re: Help Me ! » Extremely complicated change of a volume formula » 2005-12-27 09:48:53

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

#6 Re: Help Me ! » Extremely complicated change of a volume formula » 2005-12-26 09:09:36

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

#7 Re: Help Me ! » Extremely complicated change of a volume formula » 2005-12-25 12:31:57

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.

#8 Help Me ! » Extremely complicated change of a volume formula » 2005-12-25 10:12:27

fizzled
Replies: 25

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

Board footer

Powered by FluxBB