Math question - calculating mass
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Nearoth [ Major Voice ]   Joined: August 10, 2004
Posts: 395
Submissions: 76
Location: Holland

 Math question - calculating mass Posted on Thu Feb 27, 2014 1:18 pm || Last edited by Nearoth on Fri Feb 28, 2014 2:57 pm; edited 1 time in total Greetings all, I've got a math problem. I'm trying to make a formula to calculate how much material I need for a certain weave using a certain gauge. I've measured the weight of certain weaves using certain materials of a certain thickness. What I'm trying to do if have a formula to calculate how much mass a certain weave would have if I would change the wire thickness, then I can use that number to calculate how much material I would need (useful for someone ordering a weave of a certain wire thickness in silver or other precious metal) I've got this formula to calculate the mass of a ring; ID = wire thickness * AR wire length = 2*pi*(0,5 ID+wire thickness) = 2*pi*(0,5*wire thickness*AR+wire thickness) mass = pi*0,5*wire thickness^2*wire length Hope someone can help me figure this out, my head is starting to spin Thanks in advance.

Joined: July 25, 2013
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Location: Huntington, Indiana, USA Posted on Thu Feb 27, 2014 2:29 pm AR = ID/Thickness ID = AR*Thickness Wire Length = 2*pi*radius Wire Length = 2*pi*((ID+Thickness)/2) = pi*(ID+Thickness) = pi*(AR*Thickness+Thickness) = pi*(AR+1)*Thickness Mass = Wire Length * Thickness * Density Mass = (AR+1)*pi*Density*Thickness^2 You have to know the density of the material you're using for your rings to get the actual mass. Alternatively, you could weigh a known number of rings and take that weight, divide by the number of rings, plug that in for the Mass, and determine the density for future reference. -Lucidish Closing rings since 2013-07-27 A jack of all trades, master of none, often times better than a master of one. Maille Code V2.0 T4.2 R4.1 En.o Fper MFe.s W\$ C\$ G0.8-2.0 I3.2-9.5 N8.8 Pjt Djt X0 S13 Hn
Nearoth [ Major Voice ]   Joined: August 10, 2004
Posts: 395
Submissions: 76
Location: Holland Posted on Thu Feb 27, 2014 6:12 pm Thanks Lucidish, I know the density or mass number (not shure what it's called in English) of the different materials. The reason I added the wire size to the inner dimension is to get to the centerline of the wire. The inner dimension is on the outside of the wire. I was trying to eliminate all the variables in the formula except for the wire thickness. Since the AR would remain the same when changing the weave. I've calculated from several weaves the weight/cm (for a set wire thickness). I know how to change from one material to the next with the index I made but not from one thickness to the next. I'm having trouble with the formula because not only does the diameter of the wire change but also the length (due to the inner dimension changing with the set AR) the length of the wire (circle) would be; 2*pi*radius = 2*pi*1/2ID Since I want the center of the wire I take half the wire thickness on each side; 2*pi*(1/2ID+2*1/2wire thickness) = 2*pi*(1/2ID+wire thickness) For the mass I take the surface of the wire times the length; pi*radius^2*wire length = (pi*1/2wire thickness^2)*(2*pi*(1/2ID+wire thickness)) ID = AR*wire thickness (pi*1/2wire thickness^2)*(2*pi*(1/2(AR*wire thickness)+wire thickness)) When I try to simplify that I get stuck... (pi*1/2wire thickness^2)*(2*pi*(1/2(AR*wire thickness)+wire thickness) = (1/2pi*wire thickness^2)*(pi*AR*wire thickness)+wire thickness = (1/2pi*wire thickness^2*pi*AR*wire thickness)+wire thickness = (1/2pi*pi*wire thickness^2*wire thickness*AR)+wire thickness = (1/2pi^2*wire thickness^3*AR)+wire thickness = Still with me? I got different formulas every time I tried it, not sure how... Is this correct?

Joined: February 15, 2002
Posts: 393
Submissions: 10 Posted on Thu Feb 27, 2014 6:35 pm Let's make it simple. You have a ring of a certain size, and it takes n rings to make some size patch. You have another ring that is half the size of your first ring. This ring has 1/8 the mass of the first ring. I'm not going to go into the math of a toroid, but instead the easy math of a cube. [Volume 1] : x * y *z [Volume 2 (half size)] = (x/2) * (y/2) * (z/2) = x * y * z / 8 = [Volume 1] / 8 All 3-D volumes work this way, no matter what their shape. For the area, you used n rings for your first patch. With half-sized rings, you'll need 4 * n rings. Again, let's go with the easy math of a square: [Area 1]: x * y [Area 2] = (x/2) * (y/2) = x * y / 4 = [Area 1] / 4 So, when using a half-sized ring, you'll need 4 times as many rings, but each ring weighs 1/8 as much as your original ring. 4 * 1/8 = 1/2, or the ratio of your new wire diameter to the original wire diameter. Yes, it really does work out to being that simple. *** NOTE: THIS IS ONLY TRUE IF YOU ARE NOT CHANGING YOUR ASPECT RATIO!!! *** IGP (Irregular Grid Painter) Links: Home | FAQ | Downloads
Nearoth [ Major Voice ]   Joined: August 10, 2004
Posts: 395
Submissions: 76
Location: Holland Posted on Fri Feb 28, 2014 11:41 am It feels weird it could be that simple... I was hoping that by simplifying the formulas (also because the AR was a constant) it would become a simple formula like that. I guess I could get there if my math was better I currently only use this to make chains (necklaces and the like). So am I right to understand that is I would half the wire size this would make me use 2 times n rings. Length just being x = (x/2) and the volume still be (x*y*z)/8? So this would mean that when I want to calculate this with a not perfect number (like when the size if halved). I use for the Length [Length]: (x / ([original wire thickness]/[new wire thickness])) for halving this would look like: [Length]: (x / (2/1)) = (x/2) And for the volume: [Volume]: (x / ([original wire thickness]/[new wire thickness])) * (y / ([original wire thickness]/[new wire thickness])) * (z / ([original wire thickness]/[new wire thickness])) for halving this would look like: [Volume]: (x / (2/1)) * (y / (2/1)) * (z / (2/1)) = (x/2) * (y/2) *(z/2) * I hope I haven't overcomplicated it again... The AR will remain largely the same. In the end it only is used for an estimate (mostly for precious metals) and if there would be someone who would be interested in a different metal I don't usually use, like gold.

Joined: February 15, 2002
Posts: 393
Submissions: 10 Posted on Fri Feb 28, 2014 5:31 pm You've pretty much got it. Since you are figuring out a chain, you will only need twice as many half-sized rings (as you calculated above), and your half-sized rings still weigh 1/8 as much as your original rings. 2 * 1/8 = 1/4, which is the ratio of your new wire diameter to the original wire diameter, squared. IGP (Irregular Grid Painter) Links: Home | FAQ | Downloads

Joined: March 3, 2002
Posts: 990
Submissions: 244 Posted on Fri Feb 28, 2014 7:14 pm Yep, it's pretty easy as long as you don't change the AR at all. If you do that then you'll need some serious geo-trig and calculus or a huge sample set to get accurate numbers. The main thing to be careful of is the accuracy of your measured inputs like weight, number of rings and AR. Even rounding errors will be multiplied by the size of your project and can cause a really significant problem with larger projects in precious metal. www.mailletec.com Y'know, that might just be crazy enough to work!
Nearoth [ Major Voice ]   Joined: August 10, 2004
Posts: 395
Submissions: 76
Location: Holland Posted on Fri Feb 28, 2014 8:17 pm Question: I would like to calculate the volume/mass for the same length of chain. I know that for instance my JPL made with 0,8mm wire weighs 0,34 grams per centimeter When I want to know what the same AR of 1mm wire would weigh for the same centimeter So I'd use the same amount of rings the volume would increase by; for x [New] / [Original] * for y [New] / [Original] * for z [New] / [Original] for x ( 1/0,8 ) * for y ( 1/0,8 ) * for z ( 1/0,8 ) 1,25 * 1,25 * 1,25 = 1,953125 = 1 61/64 The total length would increase by; for x [New] / [Original] for x ( 1/0,8 ) = 1,25 So if the new volume would be 1,953125 times the original that would be for 1,25 times the original size. To get the volume for the original length you would again divide it by 1,25? Could you also just multiply the volume by Y and Z because you want the X to remain the same? Or am I going up something... Again this is an example. I want to convert all the number of the weaves I have to 1mm wire so it's easier later to change it with excel formula's. (I'm using more rounded number in the examples above) Again thanks for all your input.  