The Picard Maneuver@lemmy.worldM to Lemmy Shitpost@lemmy.world · 22 hours agoThe grand prizelemmy.worldimagemessage-square119fedilinkarrow-up1809arrow-down16
arrow-up1803arrow-down1imageThe grand prizelemmy.worldThe Picard Maneuver@lemmy.worldM to Lemmy Shitpost@lemmy.world · 22 hours agomessage-square119fedilink
minus-squareweker01@sh.itjust.workslinkfedilinkarrow-up3·edit-213 hours agoThat is… surprisingly little. Are you sure?
minus-squareJiggle_Physics@lemmy.worldlinkfedilinkarrow-up2·9 hours agoI mean, I can only estimate it’s size from the person standing next to it. From there I can use that estimate to get the volume of the cube, then the weight, then look up the cost by weight right now and apply the average. So it would be somewhere around 1mm by weight.
minus-squareAlwaysnownevernotme@lemmy.worldlinkfedilinkarrow-up1·12 hours agoBy weight probably, for it to be a perfectly symmetrical cube would likely cost you double that.
That is… surprisingly little. Are you sure?
I mean, I can only estimate it’s size from the person standing next to it. From there I can use that estimate to get the volume of the cube, then the weight, then look up the cost by weight right now and apply the average.
So it would be somewhere around 1mm by weight.
By weight probably, for it to be a perfectly symmetrical cube would likely cost you double that.