Torpedo practice 1956(?) camera Canon VT Deluxe, shutter 1 millisecond, lens 50 or perhaps 100 mm, film the fastest Kodachrome the Midway NX sold. The camera was right-side-up (RF on top). Torpedo was doing about 50 knots; note the black-powder-launching charge smoke. My question is "Is the torpedo longer or shorter than shown?" Bob Youngblood
Longer, it hasn't cleared the launch tube yet.
I saw the torpedos on the submarine I was on and the nose color was about 1/5 the length of the torpedo as I remember.
I was asking whether the visible portion of the torpedo has been stretched or shrunk by the moving shutter-slit in the camera. I don't recall which way the Canon slit went; in my antique Leica, it moves right to left over the inverted projected image. That's why I specified that the camera was right-side-up. I know how big the tin-fish was; I occasionally helped load the tubes.
I was very pleased to finally catch that shot; I must have wasted a dozen frames during previous practices, shooting from my battle station, the gun director. Bob Youngblood
It's Tuesday (5/21/1645e now); I haven't seen today's Daily Hedgehog nor any other replies. runnyblood
Did I get any serious answers? Or does nobody remember focal-plane shutters? Runnyblood
Solving it myself?: assume shutter direction and speed right-left, 1/20 second (per my ancient Leica).
Torpedo speed est'd 50 kts~60 mph~1 mile/min~90'/second~4.5ft in 1/20 second. Torpedo is 1/5 of image.
Image is inverted; slit will pass nose first, launcher 1/100 second later; fish moved~1 foot, so the visible part is a foot LONGER than the image indicates, and fully airborne by end-of-frame (now recording the left half of Mount 42 quad-40 antiaircraft gun). Did I do that right? Runnyblood
If you want to reply, then
register here. Registration is free and your account is created instantly, so you can post right away.