- Oct 12, 2022
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Revisiting the lunar landing site visibility question, recalled I used tangents to come to an estimate. But then remembered that slide rules are limited when it comes to small angles. It’s been a while since I picked up a book of logarithms and can’t recall the limits there. However, recalled that if an angle is in radians, the arc length = radius x angle in radians. An arc of a radius of 233,816 miles is close enough to flat over the 14 foot width of the LEM bottom stage not to matter, so 14 / 1,234,548,480 gives 1.134x10^-8 radians. The human eye can, according to one search today, resolve down to 1/60 of a degree, which is 2.909x10^-4 radians. Dividing that by about how wide the LEM bottom stage would appear from earth gives 25,652.557 power as what’s needed to see it. Without checking, pretty sure that’s different from what I came up with before.
Main point was to see if I could side-step the limit of a slide rule. No, didn’t use one for this. Still, it could be calculated on one, and might try it later.
Why a slide rule? Because the Apollo astronauts carried one just in case. It turns out it was a little 5 inch slide rule, the Pickett N600-ES.
Okay, so all that above is useless, but stuff like that bugs me.
Main point was to see if I could side-step the limit of a slide rule. No, didn’t use one for this. Still, it could be calculated on one, and might try it later.
Why a slide rule? Because the Apollo astronauts carried one just in case. It turns out it was a little 5 inch slide rule, the Pickett N600-ES.
Okay, so all that above is useless, but stuff like that bugs me.
