▲ 636 ▼ There was an attempt at scale. (mander.xyz) submitted 6 months ago by fossilesque@mander.xyz to c/science_memes@mander.xyz 69 comments fedilink hide all child comments
[–] Asafum@lemmy.world 139 points 6 months ago (9 children) 30,000 is roughly 1/3 of 60,000,000. VERY roughly. Lol permalink fedilink source hideshow 9 child comments replies: [–] DiarrheaSommelier@lemmy.ca 78 points 6 months ago (5 children) Logarithmically scaled image. I'll leave the determination of the base of the Log as an exercise for the viewer. permalink fedilink source parent hideshow 5 child comments replies: [–] gmtom@lemmy.world 21 points 6 months ago (1 child) I would show my proof, but I don't have enough space in this margin permalink fedilink source parent hideshow 1 child comment replies: [–] Jamesian@sh.itjust.works 4 points 6 months ago I'm here for this comment all day. permalink fedilink source parent [–] neatchee@piefed.social 4 points 6 months ago* well, we know bison in the middle are worth approximately 75 each... permalink fedilink source parent [–] davidagain@lemmy.world 2 points 6 months ago* (last edited 6 months ago) That's what I thought, so I investigated. The base of the log can be accounted for by a constant scale factor, because, for example, if n is the number of bison, log10(n) = log10(e^ln(n)) = ln(n) log10(e) and log10(e) is a constant. This change of base is a linear scale on the logs. Hence we can just take log 10 of the numbers of bison, and scale the answer by a constant factor which is log10(correct base), getting 7.778, 2.477 and 4.477 Scale that by about 2 = log10(100) to match the 5 bison in the middle pictogram, and there should be 16, 5, 9 bison on a logarithmic scale. The diagram is also wrong if it's logarithmic. permalink fedilink source parent [–] ShotDonkey@lemmy.world 1 point 6 months ago OBVIOUSLY!! permalink fedilink source parent [–] AppleTea@lemmy.zip 11 points 6 months ago yeah this graphic is terrible permalink fedilink source parent [–] muhyb@programming.dev 6 points 6 months ago Yeah, we need 799613 more bison images to justify the graphic. permalink fedilink source parent [–] Objection@lemmy.ml 3 points 6 months ago It's only off by roughly 20,000,000 permalink fedilink source parent
[–] DiarrheaSommelier@lemmy.ca 78 points 6 months ago (5 children) Logarithmically scaled image. I'll leave the determination of the base of the Log as an exercise for the viewer. permalink fedilink source parent hideshow 5 child comments replies: [–] gmtom@lemmy.world 21 points 6 months ago (1 child) I would show my proof, but I don't have enough space in this margin permalink fedilink source parent hideshow 1 child comment replies: [–] Jamesian@sh.itjust.works 4 points 6 months ago I'm here for this comment all day. permalink fedilink source parent [–] neatchee@piefed.social 4 points 6 months ago* well, we know bison in the middle are worth approximately 75 each... permalink fedilink source parent [–] davidagain@lemmy.world 2 points 6 months ago* (last edited 6 months ago) That's what I thought, so I investigated. The base of the log can be accounted for by a constant scale factor, because, for example, if n is the number of bison, log10(n) = log10(e^ln(n)) = ln(n) log10(e) and log10(e) is a constant. This change of base is a linear scale on the logs. Hence we can just take log 10 of the numbers of bison, and scale the answer by a constant factor which is log10(correct base), getting 7.778, 2.477 and 4.477 Scale that by about 2 = log10(100) to match the 5 bison in the middle pictogram, and there should be 16, 5, 9 bison on a logarithmic scale. The diagram is also wrong if it's logarithmic. permalink fedilink source parent [–] ShotDonkey@lemmy.world 1 point 6 months ago OBVIOUSLY!! permalink fedilink source parent
[–] gmtom@lemmy.world 21 points 6 months ago (1 child) I would show my proof, but I don't have enough space in this margin permalink fedilink source parent hideshow 1 child comment replies: [–] Jamesian@sh.itjust.works 4 points 6 months ago I'm here for this comment all day. permalink fedilink source parent
[–] Jamesian@sh.itjust.works 4 points 6 months ago I'm here for this comment all day. permalink fedilink source parent
[–] neatchee@piefed.social 4 points 6 months ago* well, we know bison in the middle are worth approximately 75 each... permalink fedilink source parent
[–] davidagain@lemmy.world 2 points 6 months ago* (last edited 6 months ago) That's what I thought, so I investigated. The base of the log can be accounted for by a constant scale factor, because, for example, if n is the number of bison, log10(n) = log10(e^ln(n)) = ln(n) log10(e) and log10(e) is a constant. This change of base is a linear scale on the logs. Hence we can just take log 10 of the numbers of bison, and scale the answer by a constant factor which is log10(correct base), getting 7.778, 2.477 and 4.477 Scale that by about 2 = log10(100) to match the 5 bison in the middle pictogram, and there should be 16, 5, 9 bison on a logarithmic scale. The diagram is also wrong if it's logarithmic. permalink fedilink source parent
[–] AppleTea@lemmy.zip 11 points 6 months ago yeah this graphic is terrible permalink fedilink source parent
[–] muhyb@programming.dev 6 points 6 months ago Yeah, we need 799613 more bison images to justify the graphic. permalink fedilink source parent
[–] Objection@lemmy.ml 3 points 6 months ago It's only off by roughly 20,000,000 permalink fedilink source parent