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Post by homobibiens on Aug 9, 2010 17:58:38 GMT -4
Well I can see that gravity is a constant downward acceleration, but isn't the initial upward acceleration not constant? Once the boot leaves the ground, there is no longer any upward accelertation. And wouldn't that mean that the upward trip might be of shorter duration than the downward trip? Or vice versa If we measure the time up from the moment the boot leaves the ground, and the time down to the moment the boot returns to the ground, and the astronaut remains in a perfectly rigid shape and orientation during the trip, then the times would be the same, because while the astronaut is not in contact with the moon's surface, there is no force but gravity acting on him. If he changes shape (stretches out or bunches up while off the ground), or spins (takes off vertically and lands horizontally, for example), all bets are off, because the relation between the boot and the center of gravity would be changing.
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Jason
Pluto
May all your hits be crits
Posts: 5,579
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Post by Jason on Aug 9, 2010 17:59:49 GMT -4
Can I just clarify something, I don't think it comes down at the same speed. It comes down from it's highest point in the same time it took to reach it........ Bob B can clarify this? That was my point. I don't think the trip up is necessarily the same duration as the trip down.
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Post by homobibiens on Aug 9, 2010 18:17:33 GMT -4
I believe the Mythbusters busted that one too. Bullets fired straight up don't land with lethal speed, while if they are fired at an angel so they form a parabolic flight path then they can in fact still kill people. I am getting that (in a vacuum) the longest flight path would be for a bullet fired up at about 56 degrees. So if the atmosphere were very thin, and didn't affect the trajectory much, a bullet fired at this angle would be slowed down the most. I'll see if I can work out a more realistic analysis a bit later.
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Bob B.
Bob the Excel Guru?
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Post by Bob B. on Aug 9, 2010 18:48:51 GMT -4
Let’s see if I can help clarify all the questions etc. regarding the flight of a projectile.
Case 1 – On the Moon
The Moon has no atmosphere; therefore drag is not a factor. When a body is thrown, the only force acting on it is gravity. The trajectory/velocity/time of the body on the way down is exactly the same as it was on the way up only in reverse.
Case 2 – On Earth
The Earth has an atmosphere; therefore drag is a major factor. Gravity acts vertically downward while drag acts in the direction opposite the velocity vector. If a body is thrown straight up, both gravity and drag act downward as the body rises; however, on the way down, gravity acts downward but now drag acts upward. The body is decelerated more quickly on the way up than it is accelerated on the way down. Because of this, the body is traveling slower when it returns to the ground than the velocity it was given initially. Because the average speed on the way up is greater than on the way down, it takes more time to come down than it did to go up. The same it true if the object is throw at an angle, but the geometry is a bit more complex.
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Post by frenat on Aug 9, 2010 18:54:14 GMT -4
Thanks, Bob. I figured air resistance played a role in how fast the bullet would fall, but I thought free falling objects have a maximum speed determined by the amount of gravity. Only the speed of light.
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Bob B.
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Post by Bob B. on Aug 9, 2010 19:16:19 GMT -4
Thanks, Bob. I figured air resistance played a role in how fast the bullet would fall, but I thought free falling objects have a maximum speed determined by the amount of gravity. You may be thinking of terminal velocity. Terminal velocity occurs when the gravity trying to accelerate a free-falling body is exactly balanced by the drag acting in the opposite direction. The body will then fall at constant velocity (though not really because the steadily thickening air at lower altitude causes a body’s terminal velocity to constantly decrease as it falls). However, a body can travel faster than terminal velocity if it is given an initial velocity that is greater than its terminal velocity. In this case, drag will simply slow the body down until terminal velocity is reached. In the case of a bullet fired into the air, it will decelerated very quickly at first because its high velocity produces a lot of drag. However, before it slows down significantly, it will cover a large distance. Eventually drag and gravity will slow it to zero vertical velocity and its will start to fall back toward the ground. On the way down, however, it will simply speed up until it reaches terminal velocity and then freefall back to the ground. It will take a lot longer to cover the distance coming back down than it did going up. Perhaps counter intuitively; a bullet fired straight down toward the ground from high in the air will also hit the ground traveling at terminal velocity if fired from a significantly high altitude. As soon as the bullet leaves the muzzle it will start to slow down because the drag, which is acting upward, is very much higher than the gravity pulling the bullet downward. As long as the drag is greater than gravity, the bullet will slow down, and the drag will remain higher than gravity until terminal velocity is reached.
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Post by LunarOrbit on Aug 9, 2010 19:18:29 GMT -4
Okay, I'm not doubting you, I'm just having a hard time wrapping my head around it. I just don't understand why the speed going up is in any way connected to the speed going down. Let's say I'm standing on a platform 3 feet above the Moon's surface, and then suddenly the platform disappears. How fast would I fall? Would it be the same speed as an astronaut who had jumped up 3 feet and then fell back down? We both began our fall at the same height (3 feet). We both began our fall at the same speed (0 mph). So what difference does it make how quickly the jumping astronaut got up to the 3 feet? (This is why I usually leave these kind of questions alone )
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Post by frenat on Aug 9, 2010 19:34:27 GMT -4
Okay, I'm not doubting you, I'm just having a hard time wrapping my head around it. I just don't understand why the speed going up is in any way connected to the speed going down. Assuming no air resistance (so no terminal velocity) and the starting and ending positions at the same height, then it takes gravity the same amount of time and distance to decellerate the astronaut (or any other object) on the way up as it has to accellerate on the way down. Ending speed should be the same as the starting speed and time of both up and down trips are the same. Let's say I'm standing on a platform 3 feet above the Moon's surface, and then suddenly the platform disappears. How fast would I fall? Would it be the same speed as an astronaut who had jumped up 3 feet and then fell back down? Yes. We both began our fall at the same height (3 feet). We both began our fall at the same speed (0 mph). So what difference does it make how quickly the jumping astronaut got up to the 3 feet? It doesn't. The trip up for the jumping astronaut will take the same amount of time as the trip down though.
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Bob B.
Bob the Excel Guru?
Posts: 3,072
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Post by Bob B. on Aug 9, 2010 19:50:36 GMT -4
Let's say I'm standing on a platform 3 feet above the Moon's surface, and then suddenly the platform disappears. How fast would I fall? Would it be the same speed as an astronaut who had jumped up 3 feet and then fell back down? Yes, you would fall at the same speed as the astronaut. We both began our fall at the same height (3 feet). We both began our fall at the same speed (0 mph). So what difference does it make how quickly the jumping astronaut got up to the 3 feet? For the trip back down, it doesn’t matter how quickly the astronaut got up to 3 feet. What I’m saying is that, for an object in free flight, the time it takes to go from zero to 3 feet is the same as it takes to go from 3 feet to zero (neglecting all the complicating issues related to how the astronaut pushed off, etc.). I just don't understand why the speed going up is in any way connected to the speed going down. In the absence of an atmosphere, the only force acting on an object in free flight is gravity. Therefore, an object traveling upward is decelerating by gravity and an object traveling downward is accelerated by the same gravity. On the Moon the acceleration of gravity is 1.623 m/s 2. If you give an object an initial upward velocity of 1.623 m/s, it will slow to zero velocity is exactly one second. During that one second it will travel a distance of 0.8115 meter (the average velocity multiplied by time). When it reaches its apex, it will immediately start to accelerate downward at the same 1.623 m/s 2. The object will travel the same 0.8115 meter in exactly one second and return to its starting point traveling 1.623 m/s. The time it takes to fall is not dependent on the time it takes to rise, but if the only force acting on the body is gravity for both the upward and downward journeys, the time will be the same.
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Post by theteacher on Aug 9, 2010 20:14:09 GMT -4
Okay, I'm not doubting you, I'm just having a hard time wrapping my head around it. I just don't understand why the speed going up is in any way connected to the speed going down. Maybe it is clarifying to consider the phenomenon in acceleration terms. On the way up, the body is slowed down by the acceleration of gravity until the speed is reduced to zero. Now the fall begins with exactly the same acceleration but in the opposite direction, so therefore the body will reach the ground (or the starting level) in exactly the same time it took to reach the apex. Yes, it would. Or more correct: It would take the same time to reach the ground. It doesn't make any difference how he got there what the acceleration downward is concerned, because the two astronauts have the same potential energy as they are at the same height (m x g x h). BUT if the astronaut has to JUMP up to that height, he has to start out with exactly the same speed, that he will reach as he hits the ground after jumping down, ie the kinetic energy when he starts the free flight phase of the jump (after leaving the ground) must be the same as it is when he reaches the ground again (½ x m x v^2), which follows from the law of conservation of energy. Edited to add: So if we look at the math: ½ x m x v^2 = ½ x m x v^2 and we consider the mass to be unaltered, the speed v has to be the same on both sides of the equation to make it hold true.
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Post by dwight on Aug 9, 2010 20:54:30 GMT -4
BTW what is a "time click"? I ask because the RCA TV camera was a field-sequential camera operating at 30fps (or 60 fields), Due to the nature of sequential color matrixing one sequential color field was always delayed so as to make it compliant with the other two color fields. That would play havoc with accurate frame analysis because you would never really know what particular moment you were looking at- at least not without the raw sequential tape. For fluid video it is not really an issue, but for field/frame analysis it becomes a nightmare. Of course, that, along with MPEG2 to FLV/RM issues were something you factored in to your anaylsis, correct Rodin?
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Post by gillianren on Aug 9, 2010 21:42:06 GMT -4
Who was it said - by their works ye shall know them? And we know you haven't done any yet. So.
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Post by homobibiens on Aug 9, 2010 21:51:46 GMT -4
Okay, I'm not doubting you, I'm just having a hard time wrapping my head around it. I just don't understand why the speed going up is in any way connected to the speed going down. The answers from frenat, Bob B, and theteacher look correct to me - I'm just offering an interpretation with a slightly different emphasis here. Let's say I'm standing on a platform 3 feet above the Moon's surface, and then suddenly the platform disappears. How fast would I fall? Would it be the same speed as an astronaut who had jumped up 3 feet and then fell back down? Yes, it would. It would take about 1.06249 seconds to fall this distance, and when you hit the ground, you would be traveling about 5.6471 feet per second, which is about 3.8503 miles per hour, or 6.19645 kilometers per hour. It would be exactly the same for the astronaut, provided his jump took him to a maximum height of three feet. We both began our fall at the same height (3 feet). We both began our fall at the same speed (0 mph). Correct. So what difference does it make how quickly the jumping astronaut got up to the 3 feet? In order to meet the two conditions in the previous statement (you both start at a height of three feet, with velocity of zero), the astronaut has to leave the ground with just the right speed. If he jumps too slowly, then he won't reach a height of three feet. If he jumps too quickly, he will reach a height of three feet very quickly, but will keep going to some greater height. The only way the astronaut can reach a maximum height of three feet is to leave the ground with a speed of 5.6471 feet per second. If he jumps more slowly than this, he won't reach a height of three feet at all. If he jumps more quickly, he will reach a greater height. If he jumps at exactly this speed, he will reach a height of exactly three feet, and will do so in 1.06249 seconds. Note that if he jumps too quickly, he will go up three feet in less than 1.06249 seconds, and will cover the last three feet on the way down in the same amount of time, which is faster than you from the disappearing platform. This is because by the time he got back down to three feet, he had some downward velocity. Hope that helps. Again, the other posts look correct, this is just a different perspective.
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Post by echnaton on Aug 9, 2010 22:31:27 GMT -4
That is not an analysis, just a bunch of images stitched together. What I am asking for is explanations and calculations that make sense of the images and logically lead to a conclusion. Got anything like that?
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Post by PhantomWolf on Aug 9, 2010 23:11:54 GMT -4
Well not that it's going to be perfect, but my initial analysis on the jump by pulling it into a frame animator shows that it took 19 frames from the time the feet left the ground until the bottom of the PLSS was stationary, and 19 frames from the time the PLSS started to move down, until his feet were on the ground again. In my cut down version, the jump started on frame 4, and the PLSS stops at frame 22. The PLSS starts to move again on frame 30 and he lands in frame 48. I have yet to graph each frame though, but I'll post it when I have. I suspect that it could be tricky though, as he rotates in mid-"air" meaning that it'll be hard to locate a "stationary" spot to use as a reference.
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