|
Post by chew on Jan 9, 2012 17:13:56 GMT -4
I've never used a sextant - what particular form would a "sextant contest" take? Who ever gets closest to their real position. When the frigate I was serving on got underway I would challenge my fellow quartermasters to get the closest Sun line to our GPS position. The bet was a soda.
|
|
|
Post by chew on Jan 8, 2012 12:35:50 GMT -4
I'd always thought the CMP was the 2nd senior astronaut and the LMP was the junior astronaut? Is this not true?
|
|
|
Post by chew on Jan 7, 2012 0:14:35 GMT -4
Balsa wood and tape... that's the extent of the technology NASA had to make fake photographs?
|
|
|
Post by chew on Jan 6, 2012 23:56:30 GMT -4
Dunno. Problems that show whatever it is wasn't there when the photograph was originally taken: There aren't any fiducials visible on the tape or wood. The fiducial plate was very close to the film, too close to tape something between the lens and fiducial plate, yet this object is blocking the crosshairs. The tape edges and wood edges are almost in focus but probably should be a lot more out of focus considering its proximity to the lens. For its focus level there is no tape grain or wood grain visible. The shadows are wrong! The wood is not face on to the lens; if it were taped to a glass plate normal to the lens we should only see one side of the wood, the closest side, but we can see two sides even though it is almost in the center of the photograph. Magazine in question: www.lpi.usra.edu/resources/apollo/catalog/metric/revolution/?AS15R15
|
|
|
Post by chew on Jan 4, 2012 17:37:28 GMT -4
|
|
|
Post by chew on Jan 2, 2012 17:34:11 GMT -4
Is that by the same guy who made this thread?I'm thinking it must be.
|
|
|
Post by chew on Jan 1, 2012 18:30:22 GMT -4
Buoyancy is definitely a factor, but it is only a second order force at the time of peak g-force, at which time water drag is far and away the major force. As the capsule slows down and plunges deeper into the water, buoyancy takes over as the first order force. Of course all of this occurs in just a matter of milliseconds. Chew, it looks like your calculations are based on the full cross-sectional area of the capsule impacting the water surface at initial contact. Yeah. I realized that after I started playing around with the numbers again. I adjusted the Cd·A product to get a 6g deceleration on 3 chutes but didn't apply them to the higher speeds.
|
|
|
Post by chew on Dec 31, 2011 11:40:57 GMT -4
I was just looking at a picture of the Apollo 15 splashdown and noted that the main parachutes have quite a few vents that may have affected its drag coefficient relative to the drogue chutes. The drogues chutes had many vents as well. I can't find a picture of the drogue chutes but this painting shows the drogues chutes: www.myspacemuseum.com/prof28.jpg
|
|
|
Post by chew on Dec 31, 2011 0:04:32 GMT -4
What really counts is the impact deceleration. How do you compute it for a CM hitting water? Compute the buoyancy force (water displacement) as a function of position after contact? Use the density of seawater vice density of air in the drag equation. For a 3 chute landing (8.66 m/s) the initial g force comes out to 8g. Close to what was experienced. For a two drogue impact the deceleration would be 260g.
|
|
|
Post by chew on Dec 30, 2011 15:47:38 GMT -4
For your drogue calculations did you include the drag caused by the CM too?
|
|
|
Post by chew on Dec 30, 2011 15:08:59 GMT -4
Might the drogue and main parachutes have different designs? Parachutes are usually not solid, they have vents in the center, or they're made from ribbons with slits between them, etc. It's very possible the designs are different, though I haven't spent much time studying the issue. I'm pretty sure the mains had a vent in them, at least that's what I recall from memory. Don't know about the drogues. Plugging in all the know quantities for drag force (mass of capsule), velocity, chute area and air density, and then calculating for the drag coefficient, I get a Cd for the main parachutes of 0.7, which sounds very low to me. I've always heard that parachutes are in the 1 to 1.5 range. The vent may likely be part of the reason for the low Cd. My calculation is also slightly flawed in that I've assumed the maximum cross-section of the parachutes are normal to the air flow. The chutes are actually angled some to the flow, so the area should be a little less than I figured, resulting in a higher Cd. My calculated impact velocity of 120 mph assumed the drogues have the same Cd as the mains. If the drogue Cd is up in the 1.4 to 1.5 range, then I think chew's 83 mph calculation is spot on. Unfortunately I'm far from being an expert on parachutes. Found the problem: the mains aren't 85 feet in diameter when in use. They are 85 feet in diameter when spread out on a floor. Their cross-section diameter when in use is about 57 feet. Plugging that into the free fall equation with a C d of 1.5 results in 8.6 m/s.
|
|
|
Post by chew on Dec 30, 2011 14:41:37 GMT -4
|
|
|
Post by chew on Dec 30, 2011 14:34:09 GMT -4
You mean the astronauts were laying flat? The capsule was suspended at an angle to cut into the water and reduce deceleration. You're right, it did hang at an angle just for that reason. But the results still weren't guaranteed. I think Apollo 12 had an especially hard landing when the CM happened to hit a wave at just the wrong angle. Yeah, it was 12. Bean got beaned (heh heh) on the forehead by a camera that broke loose from it's mount from the violent impact. He needed 6 stitches. Oops. ka9q covered this already.
|
|
|
Post by chew on Dec 29, 2011 20:11:58 GMT -4
Venus, on the other hand, has a supercritical CO 2 atmosphere so dense that you don't even need a parachute to land. The Russian landers did use a parachute in the upper atmosphere, but then cut it away and floated gently down to the surface on a small flat ring around the outside of the lander. Yeah, the first time I read that it blew my mind. "No parachute??? Wow. I guess that makes sense. That is too cool."
|
|
|
Post by chew on Dec 29, 2011 19:01:27 GMT -4
Suppose the Apollo CM came down on just its two drogue chutes? What would the terminal velocity be then? I realize this isn't a very realistic scenario, as you wouldn't discover that your main chutes weren't working until after you cut away the drogues. Wikipedia, without a reference, states the drogues deployed at 24,000 feet and slowed the capsule down to 125 mph. Assuming the 125 mph speed was obtained at 20,000 feet then the capsule on two drogues would have hit at 83 mph (37 m/s). I can't find any altitude - velocity figures for the drogues.
|
|