Math problem help

[nedL]

It's not boat related, but I wasn't going to go down into the bilge for this ;) . Can someone help me out with a math problen I have (I'm one of those mathematically challenged people who gets lost in much more that basic stuff, as you'll see).
I have a volume of area which needs to have a total specific gravity of .88.

I know that the following %'s of that area have the following fixed specific gravities.

2.959% of that area has a s.g. of 8.89
4.1% of that area has a s.g. of 1.4
20.6% of that area has a s.g. of .89

How do I calculate what the specific gravity of the remaining 72.3% of the area must be in order to achieve a total s.g. of .88?

[almeyer]

Somebody please check my math:

2.959(8.89) + 4.1(1.4) + 20.6(0.89) + 72.341(x) = 100(0.88)

Solve for x, the number you're looking for is 0.52

[Ken Hutchins]

If you can explain how you find specific gravity of an area, :confused: we can probably help with the rest of the problem. ;)

[cs]

Your math is correct, but I'm not sure about the solution. The SG of .88 is an average of the whole. You've got 4 different variables so somewhere the factor of 4 needs to figure in. I come up with this:

((2.959%*8.89)+(4.1%*1.4)+(20.6%*.89)+(42.3%*X))/4 = .88

(.5038+.723x)/4 = .88

.723x = 3.52

x = 4.8686

Chad

[John Bell]

I aussume you mean volume and not area. Areas can't have a specific gravity, as they only exist in two dimensions.

It's pretty simple to solve. Start by assuming your total volume equals 1. If the specific gravity of the whole is 0.88, then the weight of 1 unit of volume is 0.88 mass units.

It's a simple matter to multiply the %ages of total volume by their densities to get the mass of each component unit.

2.96% * 8.89 = 0.2630551
4.10% * 1.4 = 0.0574
20.60% * 0.89 = 0.18334

If 1 volume has a mass of .88, then the mass of the remaining 72.34% has to be 0.88 minus the sum of the other masses we found above.

0.88 - 0.5037 = 0.3763

Since density is mass/volume, we then divide the mass of the remainder by the remaining volume to get the density of that volume.

density = 0.3763/72.34% = 0.52

So the answer is the rest of the volume has to have a specific gravity of 0.52.

Hope this helps.

[cs]

I think you guys are right. I was thinking average weight and not total weight.

Chad

[nedL]

I was typing out additional backround information & I lost it when I went to post it, so here goes agin.

This is for an electric wire/cable design that has a requirement of floating in water with a "12% bouyancy", thus the specific gravity of .88. As this is for a cable which could have a continuous lenght in the water, the actual vaolume may not be definable (I know I could use an arbitrary length of say 100' & calculate it using volumes, but I thought it could also be done by using %'s of the total area). remember, I said my math abilities aren't anything to write home about.

Tha actual scenario is

2.959% of the total cross sectional area is the copper conductors (s.g. of 8.89)
4.1% of the total cross sectional area is the insulating material for the wires (s.g. of 1.4)
20.6% of the total cross sectional area is an inner jacketing material (s.g. of .89)
This leaves 72.3% of the cross sectional area of the finished cable (final diameter is predetermined & fixed)for a second, outer jacketing material that can be foamed to adjust the s.g.. I'm trying to figure out what s.g. that outer jacketing needs to be foamed to to result in an overall s.g. of .88

(maybe you have all come to a consensus while I've been typing.)

[ken mcclure]

The amount, or length, of cable in the water is not relevant. It either all floats or none of it floats.

It would seem to me that you are more concerned with buoyancy than specific gravity. The cable itself is always going to be heavier than water, and you need enough buoyancy to overcome that difference.

[nedL]

Thanks for your help everyone. John, I completed my work as you did & came up with the answer of .52 also, but doubted myself because someone else here has a different answer.
One of my coworkers calculated it based on a 1000' length of cable, so that volumes of each component can be used. (we know the diameters of each component, and applied the 1000' length). He calculated that the outer jacketing material would need to have a s.g. of .718 to result in an overall s.g. of .88.
Thus my delema. :confused:
Maybe I'll try it his way & see what I get.

[nedL]

Ken Mc. - That's kind of the path I was heading down, thinking I could use the cross sectional area & %'s of the area.

[ken mcclure]

You might try the tack of getting the weight of a section of cable, and then seeing how much buoyancy you need to overcome it. Don't forget to include the weight of the sleeve you'll use to contain the foam.

I feel an experiment coming on .......... :eek: