Standing there under the Burnside Bridge, doing math in my head, pushing my trig tables to their limits, metering and evaluating light, thinking things through, calculating velocities of individual wedges, and finally selecting a shutter speed:
Now I got a super slick Ferris wheel.
I'm so goddamn proud of the effort it took to make this.
I could have shortcut it with multiple takes until I got it right, but no, I did the freakin' math and got it right the first time.
O.O I wouldn't even know how to approach this problem on paper, let alone in my head.
Did you know the diameter? And then count the seconds for the velocity?
This is super cool, but I def would've taken the trial and error approach:D
The diameter is d=1! And the rotation speed is counted in seconds from a marked-out gondola with an identifying feature as it passes under the outgoing strut! From there, we can count the number of gondolas n and we can determine that d/n is almost our shutter speed, but the question is, how big of a border do we want between the wedges of 2pi/n, and we can subtract that from the previously counted shutter speed!
So we don't actually care about the velocity in terms of distance because we just need to know how far one column of lights has traveled in terms of the amount of said columns? Which ofc also is a distance but different...
Basically were just looking at angular velocity in terms of angle between gondolas? And that's our shutter speed because we want one wedge per gondola.
And then to get some space in between, we subtract a little shutter speed? In terms of a full circle divided by the amount of wedges/columns/gondolas?
Ugh, I wish my math professors could come up with problems this cool.
This is super impressive, I like thus picture a lot. Not sure where I would hang it, but I'd love to have a print of this.
You got it! You’re smarter than you give yourself credit for, go easy on yourself a bit from now.