Is my idea for the number of individuals in Destoroyah accurate?

[GodzillaLouise]

In my Godzilla/Ultraman fanfic world’s fan series Ultraman M/L/V, the iteration of Destoroyah included differs vastly from the one included in Godzilla vs Destoroyah from 1995. However, one retained feature is that Destoroyah is not one organism, but rather a colony of countless microscopic crustaceans. In my version, these crustaceans are referred to as “subquantum prawn”, and are much smaller than even the smallest of subatomic particles (which is a matter that I intend to talk about separately, as I have no idea how that’s scientifically explainable): an individual subquantum prawn has a mass of 5*10^-4375 quectograms. Despite this, the subquantum prawn eventually combine to take on the form of Destoroyah’s “Completed” form: a 145 meter tall bipedal crustacean monstrosity with a weight of 750,000 metric tons.

My current idea for the number of subquantum prawn in Destoroyah’s earlier “Aggregate” form are listed on the profile document as “2.5*10^296500 to 5*10^299750”. For the record, the completed form is the combination of 15 copies of the aggregate form. Either way, I have absolutely no idea if those are at all accurate numbers, as I don’t know of any calculators that can handle such insane numbers I could use to calculate it for myself.

[GodzillaLouise]

Ok, so seeing as literally no-one’s responded in almost 12 hours, I figured I’d make a change to make it easier to calculate, for myself or others: the subquantum prawns now have a mass of 5*10^-438 quectograms, which is a far cry from the much smaller 5*10^-4375 quectograms and might actually be calculable.

[sangi39]

Ok, so seeing as literally no-one’s responded in almost 12 hours, I figured I’d make a change to make it easier to calculate, for myself or others: the subquantum prawns now have a mass of 5*10^-438 quectograms, which is a far cry from the much smaller 5*10^-4375 quectograms and might actually be calculable.

So going by this latest mass, you'd take 7.5*10^11 grams (the mass of the completed form) and divide it by 5*10^-468 grams (the mass of a single "subquantum prawn"), and you'd get 1.5*10^479 subquantum prawns per completed form (or 1*10^478 per aggregate form)

Using the original "subquantum prawn" mass, the number in a completed for would be 1.5*10^4416 (I think)

WolframAlpha handles calculations of that size pretty well, but the easiest thing to do first is just to convert all of your mass measurements into the same units (so just kilograms or grams instead of quectograms and metric tons), and then divide the larger mass by the smaller one to work out the total number of subquantum prawns you need


That mass of the individual subquantum prawn really is tiny. The electron neutrino, for example, has a mass of 1.25×10^-34 grams (from what I can find), so each subquantum prawn has a mass 2.5*10^433 time smaller than that (using the larger mass given in your second post). I'm not sure that is explainable scientifically, to be fair, at least not in terms of our current understanding of physics

[lurker]

I had no idea until this post that there were two new metric prefixes added to either end of the scale in 2022. Now the IEC needs to add robi- and quebi- as binary prefixes.

[sangi39]

I had no idea until this post that there were two new metric prefixes added to either end of the scale in 2022. Now the IEC needs to add robi- and quebi- as binary prefixes.

I tend not to use them after a point. Kilo-, mega-, and giga- for bytes, kilo- for grams, centi- and milli- for litres, and kilo-, centi-, and milli- for metres in speech, but anything outside that, especially when doing anything involving calculations, pick a set of bases, use scientific notation, and then carry on. Avoids having to memorise a set of prefixes that don't really indicate a sense of scale, and makes the maths a lot easier than "5.6x10^2 yottagrams divided by 1.25x10^-3 picograms" (although, incidentally, WolframAplha, again, can parse at least this phrase, giving the answer 4.48x10^41)