The Picard Maneuver@lemmy.worldM to Lemmy Shitpost@lemmy.world · 2 years agoThe grand prizelemmy.worldexternal-linkmessage-square137linkfedilinkarrow-up1954arrow-down16
arrow-up1948arrow-down1external-linkThe grand prizelemmy.worldThe Picard Maneuver@lemmy.worldM to Lemmy Shitpost@lemmy.world · 2 years agomessage-square137linkfedilink
minus-squareweker01@sh.itjust.workslinkfedilinkarrow-up3·2 years agoThat is… surprisingly little. Are you sure?
minus-squareJiggle_Physics@lemmy.worlddeleted by creatorlinkfedilinkarrow-up2·2 years 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-up1·2 years 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.