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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?
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Bob B.
Bob the Excel Guru?
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Post by Bob B. on Dec 30, 2011 21:29:20 GMT -4
For your drogue calculations did you include the drag caused by the CM too? No, I discounted that though I probably shouldn't have. Considering that the CM is actually bigger than the drogue, the drag on the capsule is nearly the same as that on the parachute. Recalculating, I get: CM on 2 drogue parachutes = 49.4 m/s (110 mph) CM on 1 drogue parachutes = 61.0 m/s (136 mph)
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raven
Jupiter
That ain't Earth, kiddies.
Posts: 509
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Post by raven on Dec 30, 2011 22:48:56 GMT -4
Well, such crashes are survivable, potentially, but that's going to wreak a body pretty darn badly. Getting their mangled bodies out of the door would have been a problem. I also worry about damage to the capsule from the impact causing problems, like maybe starting to sink. I am glad it never had to be tested in practise.
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Post by ka9q on Dec 30, 2011 23:23:54 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?
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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.
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Post by ka9q on Dec 31, 2011 11:11:37 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.
Also, I happened to notice that nearly all of the Apollo lunar missions splashed down along a SSW-NNE line running roughly through the Cook Islands and Hawaii. I wonder why? What were the considerations in picking any particular splashdown point?
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Bob B.
Bob the Excel Guru?
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Post by Bob B. on Dec 31, 2011 11:40:07 GMT -4
Also, I happened to notice that nearly all of the Apollo lunar missions splashed down along a SSW-NNE line running roughly through the Cook Islands and Hawaii. I wonder why? What were the considerations in picking any particular splashdown point? The return trajectories were probably timed to coincide with a particular alignment of the geomagnetic plane, thereby allowing the spacecraft to skirt along the edges of the Van Allen Radiation Belts. If so, this would make all the landings in approximately the same area. www.braeunig.us/apollo/apollo11-TLI.htm
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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
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Post by Obviousman on Dec 31, 2011 19:50:09 GMT -4
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Post by ka9q on Dec 31, 2011 21:09:55 GMT -4
I just listened to much of the Apollo 11 recovery sequence. The quality of the radio communications was even worse than I had remembered. They must have worked pretty much according to a planned script because there was no way to transfer orders with those radios.
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Bob B.
Bob the Excel Guru?
Posts: 3,072
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Post by Bob B. on Jan 1, 2012 17:55:17 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. 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. Since the capsule had a dish shaped bottom and descended at an angle, the initial contact area was smaller and gradually increased as the capsule plunged into the water. This results in the g-forces building more gradually with a lower peak. I’ve attempted to simulate the water entry as a function of time. I took into account the dish-shaped bottom but assumed a zero angle of attack to make the math easier. I don’t know how much tilting the capsule 27.5 degrees (or 29.5 degrees under the drogue) will affect the results. The geometry of the tilted scenario is way more complex than I feel like dealing with. I also figured zero horizontal velocity. Here are the results (for what they’re worth): CM on 3 main parachutes (8.66 m/s) = 6.2 g CM on 2 main parachutes (10.6 m/s) = 8.7 g CM on 1 main parachutes (15.0 m/s) = 16.1 g CM on 2 drogue parachutes (49.4 m/s) = 163 g CM on 1 drogue parachutes (61.0 m/s) = 247 g CM with no parachutes (88.5 m/s) = 519 g
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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.
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vq
Earth
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Post by vq on Jan 5, 2012 1:56:42 GMT -4
Thanks for all the work on this! 16g certainly sounds unpleasant, but the consensus seems to be that it would have been survivable under the circumstances?
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Post by zakalwe on Jan 5, 2012 6:47:22 GMT -4
Thanks for all the work on this! 16g certainly sounds unpleasant, but the consensus seems to be that it would have been survivable under the circumstances? The impact might be survivable for humans, but what about the capsule? Not much point surviving the initial impact only to drown as the cracked and crumpled capsule sank...
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Post by Vincent McConnell on Jan 8, 2012 15:34:54 GMT -4
Greetings! I have a few questions on the CM main parachute system that I hope someone here can answer. The command module used three parachutes, which Wikipedia says allowed the spacecraft to splash down at 22 miles per hour (9.8 m/s). What were the landing velocities for a single parachute failure (as occurred in Apollo 15) and a dual parachute failure (which as I understand was not survivable)? What was the terminal velocity of the spacecraft at sea level without main chutes deployed? I believe the question of what speed the capsule would hit without parachutes is answered in the film Apollo 13. I can't remember the number stated, but a news reporter contrasts the two speeds of no parachutes to parachutes.
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