y=log_a(x) <=> x=pow(a,y) where “a” is unknown. Let’s say the values of the progress bars in the image are in ]0,b]. That puts the cheetah on about b/5. It is know that cheetahs run at a maximum speed of 75mph. This gives us b/5=log_a(75) <=> pow(a,b/5)=75 <=> a=pow(75,5/b) Therefore, we have the relationship x=pow(75,pow(5/b,y)).
For the speed of light, y=((b×3)/5). It is known that the speed of light is 671000000mph. That gives us x=pow(75,pow(5/b,(b×3)/5))=671000000mph <=> pow(75,pow(5/b,b))=514285405839088.
For the airplane, y=((b×2)/5). The fastest airplane flied at about 2200mph. That gives us x=pow(75,pow(5/b,(b×2)/5))=2200mph <=> pow(75,pow(5/b,b))=227016123.
Cheetahs are 1/2 the speed of airplanes and 1/3 the speed of light. Ok, got it.
It’s possible that it’s a logarithmic scale.
y=log_a(x) <=> x=pow(a,y) where “a” is unknown. Let’s say the values of the progress bars in the image are in ]0,b]. That puts the cheetah on about b/5. It is know that cheetahs run at a maximum speed of 75mph. This gives us b/5=log_a(75) <=> pow(a,b/5)=75 <=> a=pow(75,5/b) Therefore, we have the relationship x=pow(75,pow(5/b,y)).
For the speed of light, y=((b×3)/5). It is known that the speed of light is 671000000mph. That gives us x=pow(75,pow(5/b,(b×3)/5))=671000000mph <=> pow(75,pow(5/b,b))=514285405839088. For the airplane, y=((b×2)/5). The fastest airplane flied at about 2200mph. That gives us x=pow(75,pow(5/b,(b×2)/5))=2200mph <=> pow(75,pow(5/b,b))=227016123.
514285405839088 =/= 227016123 (contradiction).
Cannot be a logarithmic scale.
it’s a double log… y = log(log(x))…
or maybe *graph not to scale
That would make way more sense when you’re talking about relative speeds of the speed of light and pretty much anything else.
They used the Windows 95 Microsoft progress bar code