Hi guys,
I need someone to double-check my calculation.
I'm trying to determine the weight of a circle of 8% tin bronze with a diameter of 1.3m and an average thickness of 0.3mm.
I'm using a density of 8800kg per cubic meter (8920 cu, 7310 sn)
I think my formula needs to be pi x radius squared x thickness x density
or
3.14159 x 0.65 x 0.65 x 0.0003 x 8800
= 3.504 kg
Does this make sense?
Sounds right to me. Area x thickness gives volume, volume x density = mass. (Note mass - not weight if you want to be picky!)
Gordon
Gordon
| Quote: |
| Does this make sense? |
Short answer: mostly. ;) Your formula is correct and plugging in those numbers gives the mass you report (3.504129486 kg).
Long answer:
First off, it would help if you could indicate significant figures. For instance, is your certainty of measurement for the copper density 8.920 * 10^3 or 8.92 * 10^3 kg/m^3? I'll assume the latter (safer), but I still have to trust that 8800 kg/m^3 is the correct density for your sample. A linear combination of your given densities of copper and tin yields [0.92 * 8920 + 0.08 * 7310] = 8791.2, which would be (8.79 * 10^3) kg/m^3 taking into account significant figures (pretending the percentages are ideal mathematical absolutes and assuming as I mentioned that each density is given to 3 sig figs). Thus "8790" would be more accurate than 8800. By the way, your densities for copper and tin are slightly different from figures I found around the web, but I'll chalk that up to purity and temperature. And I'm not a materials engineer or physicist, so I could be neglecting some property of the alloy that would make it more dense than a simple linear combination of its constituent elemental densities. Etc.
Anyway, I'm going to use 8.79 * 10^3 kg/m^3 for the density, 0.650 m for the radius, and 3 * 10^(-4) m [edit: sorry, wrote mm before] for the thickness (I'll omit the boring dimensional analysis on that one since you did the conversion correctly).
On to the calculation. You are correct in that the mass (thanks Gordon :D) will be equal to the density times the volume, and the volume in this case is the area of the circular face times the thickness of your cylinder. So "pi x radius squared x thickness x density" is correct:
3.141592654 * (0.650 m)^2 * 0.0003 m * 8.79 * 10^3 kg/m^3 = 3.50015... kg. HOWEVER, if your 0.3 mm measurement for thickness has only one significant figure (as I stated I would use), then you can only be sure of this result to one significant figure, meaning you have to round to 4 kg! If you're sure of that 0.3 mm to one more place, i.e. 0.30 mm, you can be sure it's 3.5 kg, which would be a far better result especially since you would start needing to improve multiple measurements before that result changed at all.
In other words, the above is a whole lot of verbage to say that your final mass is going to depend very sensitively on how exact you get that thickness.
Cheers,
Gabriel
PS - note that 0.3 mm would mean you're sure of the 0 and estimating the first decimal place as 3. 0.30 mm would mean you're sure of the 0.3 and are estimating the second decimal place as 0.
Last edited by Gabriel Lebec on Sat 31 Mar, 2007 9:04 am; edited 1 time in total
Two of the figures are approximations. As such the margin of error is fairly high. The intent was to approximate the mass of a Homeric shield facing. Tin content can be anywhere between 7% and 8%. The thickness is an average since the object was worked to be thicker at the centre and thinner near the edge.
The problem is that I suspect that the 0.3mm mentioned by Coles is only the thickness near the rim and, if so, I need to find a report that measures the thickness at the boss too.
The problem is that I suspect that the 0.3mm mentioned by Coles is only the thickness near the rim and, if so, I need to find a report that measures the thickness at the boss too.
There is a problem with units. The density is kg per cubic meter. You are calculating a volume in units of cubic millimeters.
A cubic millimeter is 1 X 10-9 cubic meters.
To correct your math, you need to multiply your formula for volume by 1 X 10-9 so that mm^3 is converted to m^3. The mass will then turn out to be 3.5 X 10-9 kg instead of 3.5 kg (on the order of a gram.)
A cubic millimeter is 1 X 10-9 cubic meters.
To correct your math, you need to multiply your formula for volume by 1 X 10-9 so that mm^3 is converted to m^3. The mass will then turn out to be 3.5 X 10-9 kg instead of 3.5 kg (on the order of a gram.)
| Jared Smith wrote: |
| There is a problem with units. The density is kg per cubic meter. You are calculating a volume in units of cubic millimeters.
A cubic millimeter is 1 X 10-9 cubic meters. To correct your math, you need to multiply your formula for volume by 1 X 10-9 so that mm^3 is converted to m^3. The mass will then turn out to be 3.5 X 10-9 kg instead of 3.5 kg (on the order of a gram.) |
No, he converted the mm into meters. That's what the 0.0003 is all about. The units are correct.
Edit: the equation, result, and units are all correct, but I accidentally wrote 3 * 10^(-4) mm before (instead of m). It was written correctly in the equation, but still, I apologize for any confusion that may have caused.
| Gabriel Lebec wrote: |
| No, he converted the mm into meters. That's what the 0.0003 is all about. The units are correct. |
Additionally, 3.5 x 10^-9 kg is not on the order of a gram but of a microgram... According to this page, this is about the mass of a small grain of sand. The calculations are indeed affected by the relatively low precision of the thickness, but not to that extent ;)
Regards
Yes I see that now! I thought he had a diameter of 1.3 mm (not 1.3 m.) If he had gotten an answer of 3.5 kg (3500 grams) but was off by 10-9, he would have a fraction of a gram (on the order of a milligram.)
Depending on what kind of accuracy is being sought (lab versus estimate), using percentages of components' densities does not actually work perfectly in all cases. I doubt anyone will care about .001% error, but when alloyed the spacing of constituents is not the same as that of pure elements. If you can imagine mixing two equal sized barrels, one filled with large beans and the other filled with fine grained rice, the result is less than 2 full barrels. The same phenomena occurs in alloying, but only to a very small degree.
Depending on what kind of accuracy is being sought (lab versus estimate), using percentages of components' densities does not actually work perfectly in all cases. I doubt anyone will care about .001% error, but when alloyed the spacing of constituents is not the same as that of pure elements. If you can imagine mixing two equal sized barrels, one filled with large beans and the other filled with fine grained rice, the result is less than 2 full barrels. The same phenomena occurs in alloying, but only to a very small degree.
| Jared Smith wrote: |
| If you can imagine mixing two equal sized barrels, one filled with large beans and the other filled with fine grained rice, the result is less than 2 full barrels. The same phenomena occurs in alloying, but only to a very small degree. |
I do 'that' twice daily. Pouring half a bucket of water into an4/5 full bucket of grain to get a búlging full bucket of soaked grain the next day :lol:
Mixing a variety of grains, bix, wheat chaff etc. has the same effect as alloying metals too. You end up with léss being more.
Anyway the ts had the math quite good enough for something to be used in a hobby game ;)
It is not as if it going to be a fuel thingamy in a rocket.
Thanks for the insight in the knowlegde availeable here btw. :eek:
Peter
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