This is a somewhat long and arduous question, (and is likely to help me in particular far more than it will help anyone else, since it's quite specific to my own project) but I hope you'll bear with me. I'm in the middle of writing a hard-scifi novel, and within it are massive events and battles which I intend to reflect true physics in every respect (at least relative to certain invented in-universe values). I understand many of the physical principles I need to account for and calculate, but I simply lack the math skills to calculate some of them properly.

I have a few directly interrelated problems. 1) is calculating the precise sizes and masses of various objects based on the Square Cube Law---in this case, railgun projectiles. In my novel there are 25 grades of a particular kind of railgun, each grade with its corresponding projectile:

1 | 2 | 3| 4 | 5 (1m/1.0 tons) | 6 | 7 | 8 | 9 | 10 (2m/8.0 tons) | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 | 20 (4m/64.0 tons) | 21 | 22 | 23 | 24 | 25 | [40] (8m/512 tons)

The materials, shape, and relative dimensions are identical from grade to grade. I already know that Grade-5 is 1m in diameter, with a mass of 1 ton (I simply invented that value, and used it to extrapolate the following values), Grade-10 is 2m/8tons, and Grade-20 is 4m/64tons. There is technically no Grade-40 in my novel, but as shown above, it would hypothetically be 8m/512tons. My largest grade, Gade-25, is 5m, but I haven't been able to figure out its exact tonnage. I'm sure the math is simple, but I'm ignorant of precisely what that math is, and how to accurately calculate the relative size differences of the odd in-between sizes that are not exactly double or half the size of another known value---which then leaves me unable to apply correctly modified Square-Cube values to them for tonnage.

2) Which itself is the second part of the problem. Since the in between values are not exactly double or half of the known values, I also don't know how to calculate modified Square-Cube values to multiply/divide with for those sizes (can't multiply/divide by 8 for those). It was easy for me to calculate the known values---I simply gave Grade-5 the value of 1.0 tons, and since Grade-10 is twice that size, I multiplied by 8 to get 8.0 tons, then by 8 again to get 64.0 tons for Grade-20.

3) With that being accomplished, the third related part of the problem is then calculating their kinetic energy (double the mass, kinetic energy is doubled---double the velocity, kinetic energy is quadrupled). Having now precisely calculated each projectile's mass, and assuming the same velocity for each, I need to apply the correct relative values---which I am currently unable to do for the same reasons as the above problems. To extrapolate the numbers for kinetic energy, I'll give Grade-25 a value of 1000.0 for kinetic energy, and all other grades will be proportionately less than that. Perhaps the same numbers that went into calculating the square-cube for each Grade will be used to calculate kinetic energy values for the same Grades?

Again, I know this is a long and arduous question, but if anyone could help me find the correct values for each Grade for both tonnage and kinetic energy, and show me the appropriate formulae for both (but also explain them so that a relative layperson like me can understand them, and perhaps then even be able to apply them to different situations), it would truly be most appreciated. Thanks in advance!


closed as off-topic by Aify, Vincent, sphennings, L.Dutch, Azuaron Aug 9 '17 at 13:48

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question does not appear to be about worldbuilding, within the scope defined in the help center." – Vincent, sphennings, L.Dutch, Azuaron
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  • 1
    $\begingroup$ Do you not understand how to cube a number? $\endgroup$ – A. C. A. C. Aug 8 '17 at 17:33
  • $\begingroup$ You want you Grade to be directly proportional to Diameter at grades 5, 10, 20 and 40, but want different proportion in-between? Can you try to draw a chart and tell us what do you think diameter/mass of, say, grade 8 should be? $\endgroup$ – Alexander Aug 8 '17 at 17:37
  • $\begingroup$ Yes, all are directly proportional to those diameters. My (probably incorrect) previous guess was that Grade-8 would be just under 1 meter diameter at 4.8 tons, Grade-9 would be just under 2 meters at 6.4 tons, etc. I simply don't know if those are correct values (can't remember how I came to them), and don't know how to apply Square-Cube relative to the other known values. Since Grade-10 is twice the size of Grade-5, it's easy for me to get Grade-10's mass of 8.0t. But the higher the Grade, the more increments there are in between a doubled size, which complicates the math for me. $\endgroup$ – Sci-Hard Aug 8 '17 at 18:01
  • $\begingroup$ If you want to stick with 1 ton for grade 5, your formula is simple. m = N3 / 125 (N3 is a cube of grade). $\endgroup$ – Alexander Aug 8 '17 at 18:05
  • 2
    $\begingroup$ If you have Excel, it's formula calculation probably going to be sufficient. $\endgroup$ – Alexander Aug 8 '17 at 19:43

1) The size seems to be, for instance, the diameter is (Grade/5) meters. The length would be (Grade/5) * length of grade 5 projectile.

2) This is pretty simple. It's (Grade/ 5 ) ^ 3 tons. So, for instance, a gauge 1 railgun is ( 1/5) ^ 3 = 1/125th of a ton, or 0.008 tons.

2) It's a tad annoying that you switched the standard to Grade 25, but since KE = 1/3 m x V^2, and since only m is changing, you end up with, like mass, (Grade/25)^3 x 1000.0 . If you wanted to base it off of Grade 5 instead, it would be (Grade/5)^3 x 8.0 energy units (because we pull the 1/125 out from the cube function and apply it to the constant)

  • $\begingroup$ Tell me if I did this right. To calculate KE for Grade-24, I divided 24 by 25 (0.96). Cubed=0.84934656. Multiplied by 1000=849.34656. Is this correct? $\endgroup$ – Sci-Hard Aug 8 '17 at 19:13
  • $\begingroup$ It's correct, except I have 0.96^3 = 0.884736. I'm not sure if you were cubing by hand, or entered it into a calculator incorrectly. So, I have KE = 884.736 $\endgroup$ – Jesse Aug 8 '17 at 19:18
  • $\begingroup$ I used a windows calculator. Funny though, I keep getting 0.84934656. Good to know I got the principle correct though. Thanks. $\endgroup$ – Sci-Hard Aug 8 '17 at 19:21
  • $\begingroup$ @Sci-Hard That's odd. 0.96^3 should certainly not be anywhere near 0.85. Using three different calculators (the GNOME Calculator, Perl, and my trusty old handheld electronic "scientific" calculator which has nothing to do with my PC except that it shares a desk with it as I'm trying that...) I keep getting the answer 0.884736. You are probably doing something wrong somewhere if you keep getting 0.84934656. $\endgroup$ – a CVn Aug 8 '17 at 19:49
  • $\begingroup$ @Sci-Hard 0.96^4 = 0.84934656, You cannot just square a number twice, you have to use the cube button. $\endgroup$ – Jesse Aug 8 '17 at 20:49

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