|
|
|
|
Cuneiformist
The Imperfectionist
USA
4955 Posts |
 Posted - 07/22/2008 : 08:39:00
|
Sitting in a conference this afternoon, there was brief mention of a cuneiform text referencing earthwork for a canal, and the total volume of earth dug was ca. 270 m3. Since I'd heard a version of this talk before, I zoned out and started trying to compute just what sort of cube that would be.
Some trial and error gave me a cube with sides of ca. 6.64×6.46×6.46 m.
Of course, that won't tell me much about how long such a canal would be, so I started to play around with these numbers to see what I'd come up with. Not thinking much about it (and maybe the events of previous evening's reception left me a little foggy), I just assumed that I could, say, subtract 1 m from one cube and add it to another and still not affect the total volume. But this obviously doesn't work (it quickly struck me that 4×4 is not the same as 5×3 even though 4+4 and 5+3 both equal 8).
Ultimately, I decided to make things easy and assume that the canal would be 1 m wide and 1.5 m tall. Then it was simple to find the length of such a canal (1×1.5×X=270, solving for X with 270÷1.5=180).
But this got me wondering-- is there some regular math, like a formula or something, to more quickly figure out X in that situation?
For instance, in the case of 4×4 and 5×3, I figured that in order for 5×3 to get to 16, you can do 5×3.2. But adding .2 to the smaller number obviously doesn't always work. If I have 5×5 (=25), and then 6×4.2 (both of which add up to 10), I get 25.2-- that is, .2 too much. In fact, in order to get 25 by a number multiplied by 6, you have to use 4.16 repeating.
So anyhow, back to my original question-- has anyone (at SFN or in the professional (?!) math world) thought about this?
|
|
|
Dave W.
Info Junkie
USA
26022 Posts |
Posted - 07/22/2008 : 09:32:46 [Permalink]
|
Originally posted by Cuneiformist
But this got me wondering-- is there some regular math, like a formula or something, to more quickly figure out X in that situation? |
Simpler than the single division you were doing? No.| For instance, in the case of 4×4 and 5×3, I figured that in order for 5×3 to get to 16, you can do 5×3.2. But adding .2 to the smaller number obviously doesn't always work. If I have 5×5 (=25), and then 6×4.2 (both of which add up to 10), I get 25.2-- that is, .2 too much. In fact, in order to get 25 by a number multiplied by 6, you have to use 4.16 repeating. |
Sure. If you know the length of the canal (let's say it's 27 meters), then the possible widths and heights will be W×H = (total volume divided by length). In this example, width times height must always equal 10 (so that when multiplied by length, the total volume would be 270). |
- Dave W. (Private Msg, EMail)
Evidently, I rock!
Why not question something for a change?
Visit Dave's Psoriasis Info, too. |
 |
|
|
Cuneiformist
The Imperfectionist
USA
4955 Posts |
Posted - 07/22/2008 : 10:44:12 [Permalink]
|
Originally posted by Dave W.
Originally posted by Cuneiformist
But this got me wondering-- is there some regular math, like a formula or something, to more quickly figure out X in that situation? |
Simpler than the single division you were doing? No.| For instance, in the case of 4×4 and 5×3, I figured that in order for 5×3 to get to 16, you can do 5×3.2. But adding .2 to the smaller number obviously doesn't always work. If I have 5×5 (=25), and then 6×4.2 (both of which add up to 10), I get 25.2-- that is, .2 too much. In fact, in order to get 25 by a number multiplied by 6, you have to use 4.16 repeating. |
Sure. If you know the length of the canal (let's say it's 27 meters), then the possible widths and heights will be W×H = (total volume divided by length). In this example, width times height must always equal 10 (so that when multiplied by length, the total volume would be 270). |
OK, well I was just confronted with how to manipulate 6.4×6.4×6.4 and was overwhelmed with how difficult it was to do so. If I wanted one number (e.g. the length of the canal) to go up, how could I do so so the total volume remained unchanged? It's easy enough with two numbers. But with three, it becomes harder, so I wondered if relationships, etc. had been explored... |
|
|
|
Ricky
SFN Die Hard
USA
4907 Posts |
Posted - 07/22/2008 : 12:42:43 [Permalink]
|
| If I wanted one number (e.g. the length of the canal) to go up, how could I do so so the total volume remained unchanged? It's easy enough with two numbers. But with three, it becomes harder, so I wondered if relationships, etc. had been explored... |
The common trick in 3 dimensions is to fix one number, then have a function of two:
x = 27
x*y*z = 270
z = 270/(27*y)
z = 10/y
Now you can draw out the graph of y vs. z and get a plot for all the possible values to satisfy.
In general, the information you want is stored in the derivative. But this example is so easy that it really doesn't give you anything you didn't already know. The only thing that might be useful is a linear approximation of your function at a point near (6.4, 6.4, 6.4) which would be:
f(x, y, z) ~ -524.288 + 40.96(x+y+z)
But this doesn't help much. Just gives you an easier way to find the change in the function if you're doing calculations by hand (only 1 multiplication).
|
Why continue? Because we must. Because we have the call. Because it is nobler to fight for rationality without winning than to give up in the face of continued defeats. Because whatever true progress humanity makes is through the rationality of the occasional individual and because any one individual we may win for the cause may do more for humanity than a hundred thousand who hug their superstitions to their breast.
- Isaac Asimov |
|
|
|
Cuneiformist
The Imperfectionist
USA
4955 Posts |
Posted - 07/22/2008 : 13:23:05 [Permalink]
|
Ricky, slow down. Remember: I'm in the humanities. So explain more of:The common trick in 3 dimensions is to fix one number, then have a function of two:
x = 27
x*y*z = 270
z = 270/(27*y)
z = 10/y |
Or I guess plotting the graph is all that's needed?
Mind you, I don't have any practical issue here; I was wondering if there was a formula or something for this. So perhaps this is it?
In general, the information you want is stored in the derivative. But this example is so easy that it really doesn't give you anything you didn't already know. The only thing that might be useful is a linear approximation of your function at a point near (6.4, 6.4, 6.4) which would be:
f(x, y, z) ~ -524.288 + 40.96(x+y+z) |
OK, so I don't know what you mean here. Derivative? -524/288?!? |
|
|
|
Ricky
SFN Die Hard
USA
4907 Posts |
|
|
HalfMooner
Dingaling
Philippines
15831 Posts |
Posted - 07/22/2008 : 21:45:26 [Permalink]
|
Is there any evidence that the Cuneiformians (or whatever hydrological civilization it was) did the math themselves (perhaps for the purpose of estimating the human resources needed)? Or did they just have an experienced engineer to make a guesstimate based upon his past experience?
|
“Biology is just physics that has begun to smell bad.” —HalfMooner
Here's a link to Moonscape News, and one to its Archive. |
| Edited by - HalfMooner on 07/22/2008 21:46:14 |
|
|
|
Cuneiformist
The Imperfectionist
USA
4955 Posts |
Posted - 07/23/2008 : 09:53:42 [Permalink]
|
Originally posted by HalfMooner
Is there any evidence that the Cuneiformians (or whatever hydrological civilization it was) did the math themselves (perhaps for the purpose of estimating the human resources needed)? Or did they just have an experienced engineer to make a guesstimate based upon his past experience? |
Well, the documents we have talk about things like compensating workers for labor, and it will say something like "[wages] for workers for X days of work, it (consisted of) Y [cubic units of measure] for (work on) such-and-such canal."
But that doesn't tell us quite how things were planned. There are some examples where it looks like we're seeing building plans-- or at least that suggest that Mesopotamians had that concept-- and we can imagine that earth works were similar.
So an overseer might come and figure out where he wants a canal to go, have that distance measured, and then calculate for width and depth to arrive at a total volume of earth needed to be moved. With this in mind, he could then figure out how many men he'd need (depending on how quickly the project needed to be done), and calculate the costs associated with that. |
|
|
|
Cuneiformist
The Imperfectionist
USA
4955 Posts |
Posted - 07/23/2008 : 10:02:08 [Permalink]
|
Originally posted by Ricky
No, it's not. In fact, just ignore that... When you have three variables, you have a butt-load of choices. To take away all these choices, we're going to add an additional restraint: y = z. Now our equation becomes:
x*y*y = 270. Now what happens if we change x? We get:
y^2 = 270/x
Solving for y we get:
y=sqrt(270/x)
We ignore the negative root for obvious reasons. So if we want to change x to say 20, we get that y = sqrt(270/20) = 3.674.
That is, 20*3.674*3.674 = 270. |
OK, so that makes sense. I tried to do something similar by just making z = 1, and then played with x and y. And since I knew that y was going to be a low number close to z, it was easy.
But if, for instance, you're an artist commissioned to do a piece of work, and for whatever reason you have a volume A of material, he may not want to have such constraints as y = z. (It's art!) In his case, it seems that-- as you suggested-- there's no real way to quickly figure out the dimensions of the modern art object he's going to make (assuming he's thinking of doing some rectangular object).
It would be easier for him (if knowing the dimensions were super important) to just make a prism. |
|
|
|
Ricky
SFN Die Hard
USA
4907 Posts |
|
|
Ricky
SFN Die Hard
USA
4907 Posts |
|
|
Dave W.
Info Junkie
USA
26022 Posts |
Posted - 07/23/2008 : 12:51:10 [Permalink]
|
If we restrict ourselves to even inch measurements and only rectangular shapes, and are given a cubic foot of material, we can make (not including rotations and reflections):- 1x1x1728
- 1x2x864
- 1x3x576
- 1x4x432 or 2x2x432
- 1x6x288 or 2x3x288
- 1x8x216 or 2x4x216
- 1x9x192 or 3x3x192
- 1x12x144 or 2x6x144 or 3x4x144
- 1x16x108 or 2x8x108 or 4x4x108
- 1x18x96 or 2x9x96 or 3x6x96
- 1x24x72 or 2x12x72 or 3x8x72 or 4x6x72
- 1x27x64 or 3x9x64
- 1x32x54 or 2x16x54 or 4x8x54
- 1x36x48 or 2x18x48 or 3x12x48 or 4x9x48 or 6x6x48
- 2x24x36 or 3x16x36 or 4x12x36 or 6x8x36
- 2x27x32 or 3x18x32 or 6x9x32
- 4x16x27 or 8x8x27
- 3x24x24 or 4x18x24 or 6x12x24 or 8x9x24
- 6x18x18 or 8x12x18
- 9x16x12 or 12x12x12
So that's a simple 51 different shapes (provided I've missed none and duplicated none) just from those simple rules. |
- Dave W. (Private Msg, EMail)
Evidently, I rock!
Why not question something for a change?
Visit Dave's Psoriasis Info, too. |
|
|
|
ktesibios
SFN Regular
USA
505 Posts |
Posted - 07/23/2008 : 18:44:55 [Permalink]
|
Might it be worth mentioning that a canal with a rectangular cross-section- that is, with the sides straight up and down- would be likely to cave in unless shored up?
An experienced digger would probably know enough to make the thing wider at the top than at the bottom, which would give you a trapezoidal cross-section and make your efforts to estimate the cross-section a bit more complicated.
If you know the angle of repose for the kind of soil the Mesopotamians were working with, you could estimate the appropriate ratio of top width to bottom width and take it from there. |
"The Republican agenda is to turn the United States into a third-world shithole." -P.Z.Myers |
|
|
|
tw101356
Skeptic Friend
USA
333 Posts |
Posted - 07/23/2008 : 20:14:32 [Permalink]
|
| It's easier to play around with numbers like this by breaking them down to their prime factors: 270 = 3 x 3 x 3 x 2 x 5. Then you can group them any which way. So a canal 1 meter deep and 2 meters wide would be 135 (3 x 3 x 3 x 5) meters long. Now if you wanted it 1.5 meters deep and 1 meter wide, you can break one of the 3's into 2 x 1.5, and the length would be 2 x 2 x 3 x 3 x 5 = 180 meters. |
- TW
|
|
|
|
Cuneiformist
The Imperfectionist
USA
4955 Posts |
Posted - 07/24/2008 : 03:30:34 [Permalink]
|
Originally posted by ktesibios
Might it be worth mentioning that a canal with a rectangular cross-section- that is, with the sides straight up and down- would be likely to cave in unless shored up?
An experienced digger would probably know enough to make the thing wider at the top than at the bottom, which would give you a trapezoidal cross-section and make your efforts to estimate the cross-section a bit more complicated.
If you know the angle of repose for the kind of soil the Mesopotamians were working with, you could estimate the appropriate ratio of top width to bottom width and take it from there.
|
Oh-- well you're probably right. In which case, that changes things in how to calculate. I'm sure that experience would have taught Mesopotamians some time ago that the trapezoidal shape is the superior one. But thanks for bringing that up; it's good for me to keep in mind! |
|
|
|
Cuneiformist
The Imperfectionist
USA
4955 Posts |
Posted - 07/24/2008 : 03:32:50 [Permalink]
|
Originally posted by tw101356
It's easier to play around with numbers like this by breaking them down to their prime factors: 270 = 3 x 3 x 3 x 2 x 5. Then you can group them any which way. So a canal 1 meter deep and 2 meters wide would be 135 (3 x 3 x 3 x 5) meters long. Now if you wanted it 1.5 meters deep and 1 meter wide, you can break one of the 3's into 2 x 1.5, and the length would be 2 x 2 x 3 x 3 x 5 = 180 meters.
|
Ooh. That's a quick way to start ball-parking things. |
|
|
|
 |
|
|
|
Math Q
