The input is a number that can be converted to a length. The output is a number of ore pieces. These are not the same units. There is no reason to convert anything to "cubic blocks" as a starting point when you already have a perfectly usable number given by the size. You only need two things to determine the (average) output: the shape of the curve is that of size cubed, and the coefficient. With the added results, the current average coefficient is 2.51 and with 19 samples ranges from 2.03 through 2.88 (or 2.66 if you ignore the size 3 deposits.) Given the amount of deviation seen, if you use 2.5 you are going to be as accurate as it's possible/practical to be. It's not like we're going to home in on 2.512357325723 and notice the difference (or care) that we were getting 12 more or fewer ores out of a size 10 deposit. I mean, if you're cutting it that close, you're going to get burned by the random variation anyway.

Different road, same destination. And yes, we don't need perfect accuracy. All we really need to know is that a size 10 has more ore than a size 7. lol This was a good little excercise. It's not like we are Watney trying to calculate potato yield.

Only the one size nine and it was cobalt. The edit was because I thought it was iron last night when I started mining. Woke up this morning and finished mining it out of habit. Didn't look twice at it. 'doh! Don't mine drunk kids!

Something about promethium seems to yield less for me. Size 5 - 298 Edit (didn't want to make another post since I was the last one) Mag size 6 - 577

This may have been answered somewhere else so forgive me for not researching before asking but has anyone else noticed that the yield from mining seems to be decreased if you don't grad up the ore almost right away? I am curious if it is deteriorating / disappearing or if the drill tool is actually destroying it if I hit the ore with it.

A little cheese for the numbers mice Resource: Iron Size: 8 Result: 1205 Ore Clips used: 8 Method used: circular cheesecutter, double-button mining.

Ye din add my size 15 from the very first poist to that chart ... i feel left out LOL ... And, according to that chart, the numbers for a size 15, is purdy darn close to wut i mined up

Ok, dug out some deposits last night. 5 308 6 516 7 785 7 865 8 1388 The average in all the numbers listed in this thread is now 2.503... so yeah IMO 2.5 * (size ^3) is more than close enough. 2.5 is easier to remember than 4/3 pi, and including the volume of a sphere is not a good reference IMO because very few people actually know that off the top of their head anyway. (Although I suppose most people won't be cubing numbers in their head either, so a chart will be the go-to reference anyway.) I mapped a macro to one of the side buttons on my mouse to spam the T key. I just hold that while I'm mining and then pick up the strays after a bit.

yeah.. it is looks more like 2.5*(size)^3.. but the thing is you should not get higher number of ore than the number projected by the formula..you may loose some by despawnig but never more.. there must be some more element attach to it.. the number of ore(item) dropped in the ground may be constant and the amount of ore in each of that item varies from 2-4 at random(or using some other formula)

If you roll 4d6 you wouldn't predict the result to be 24. You'd predict an average of 14 and say the range is 4-24. IMO the prediction should be an average not a maximum. My first guess would be that as vertices are moved, the amount of volume change is calculated and volume * coefficient is compared to a random number to determine if ore is dropped. But, you cannot re-generate ore by filling in the space (yes I tried, LOL) and this has to offer some clue as to how it's handled. It can't simply be "anything mined inside this sphere is ore." I think you may be right, if I'm reading you correctly. The deposit has invisible "nodes" inside it, and when the rock is mined away from them they generate an ore containing 1-4 pieces. This would prevent ore duplication. (It would still produce a curve instead of a constant maximum.)

Remember about mining speed, although it takes considerably longer, I have found that using the slow speed drill yields more units per volume, I believe its 100%, 66% and 33% for slow, fast and double speed (both buttons clicked) . However, I only have 2 size 5s of differing ore to compare. I don't think there is a difference between ore types for yield. Can anyone confirm this?

I specifically tested whether I could despawn or deterioate ore. Tested with promethium, I drilled leaving 10 units of ore on the ground. Staying on the planet and returning 2 game days later, the ore had not despawned. Using the drill tool and targeting the ore, I was not able to destroy it. The only thing I have done that I believe changes the ore yield is to use the slow speed drilling (right click) instead of full speed drilling, it seemed to me I was getting about half again as much ore.

I am not talking about predicting how many ore you get but what is the formula/mechanics attached to this ore system...the vain theory may work.. or it just generating a deposit while seeding the planet using [2.5*(size)^3] * (some random number, say from .9 to 1.1) to keep each size 6(or nay size) different from other size 6

As I mentioned on the last page, I did some tests and it didn't seem to make any significant difference: (But it does use more charge packs to mine faster, but well worth it because the time needed to mine the additional promethium used in making all the extra charge packs is way less than the time needed to mine slowly.)

Actually, cubing numbers in my head is easy - doing the volume of a sphere in my head is a little too complex even for my mutant ability. So your formula is "close enough for government work". I mapped one of my thumb buttons to the "interact" (T) function as well. Very handy for this and other purposes.