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Post by JayUtah on Nov 12, 2009 13:26:12 GMT -4
It doesn't answer my question.Of course it does. Everyone here agrees it does. But you just shift the goalposts again. Complete, succinct answers are being provided, but you are simply ignoring them and pretending you need more. Why shouldn't I conclude that you are being deliberately obtuse? I provided earlier a similar drawing to what you have just made. If that didnt answer my question, then why do you think your sketch would?Because unlike your drawing, Jason's shows the orientational relationship among the various physical and conceptual elements of a lunar orbit transfer. These elements are what I and others have been trying to tell you make it possible to design an orbit that misses most of the Van Allen belts. You are the one insisting that your one drawing must capture the whole scenario. You simply dismissed Jason's sketch outright. Hey I came here looking for answers.Hogwash. You're chock full of attitude, awarding "points" to litigants based on your limited understanding of their claims and our answers. You handwave ignorantly about orbits being straight lines. You say all scientists agree with you, but you can't name one. You dismiss as "vague" when the most eminent researcher in this field practically looks you in the eye point-blank and tells you you're wrong. Do you really think you're fooling anyone into believing you're a dispassionate, undecided observer? Or is it really that hard to do?As a drawing, yes. Jason has made a fine effort. I am an expert in 3D rendering and drawing techniques. What you ask, as a drawing, is difficult to construct in a way that represents accurately the important relationships and can be executed credibly on paper. That is why, as I said earlier, the only way to properly conceptualize these principles among us practitioners is to treat each orbit as a mathematical abstraction and rely upon the mathematics to properly describe the orientations among them. I dont have ... the knowledge to make it accurate to do it myself.Then kindly stop trying to sit in judgment on those who do. Jason has done a fine job. I know he understands these principles because I've talked with him for years. Similarly I know Bob B. is well versed in these principles because he has demonstrated his expertise in orbital mechanics. As for me, I used to design and operate spacecraft in the cislunar environment. I know what I'm talking about and I'm familiar with how one handles the cislunar radiation environment. It's time for you to realize that you cannot bluff your way through this. You're speaking to people who know what they're talking about, and you're frankly embarrassing yourself. So either somebody is up for the challenge or we have to keep using 2D representations to answer 3D questions.3D representation provided in my previous post. Unfortunately you don't get to keep persisting in your ignorance while leveling demands on others, and simultaneously implying that you and your "hoaxers" are right. If you aren't willing to make the least effort to learn something about orbital mechanics, then the obligations of others to spoon-feed you information are quickly exhausted. You get me a 3D model of what I'm looking for, and then you will have answers to your questions.No. Those are simple questions intended to discover your current state of understanding. No one is obliged to provide anything in order to discover what you think and believe. Please answer the questions. Please, enlighten me.And I feel we're about to discover that you know as little as or less about solar particle activity than you do about trapped radiation.
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Post by JayUtah on Nov 12, 2009 13:31:56 GMT -4
Jay, can I have that Van Allen belt if you're finished with it? It looks delicious. Sorry, the studio gnomes made short work of them.
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Post by fm on Nov 12, 2009 15:06:03 GMT -4
Well guys Im sooo sorry to ask you for such drawing I'm being too demanding? Lol. Come on give me a break. I thought this was a forum to debunk hoaxers. You cant possibly tell me that a 2D drawing can take everything into consideration what happened in a 3D environment. John T, curve your line! Try it, try curving your trajectory with your 2D drawing. You cant. Can you? What you are showing me that drawing is after TLI, which occurred halfway around the Earth- 1.5 orbits before injection- that the craft simply went straight? Because that's what your diagram suggests. Blast off, straight to the moon, or maybe even straight through the planet then the moon. It didnt blast off did it? It was an injection, from which I can tell means the orbit was widened from the original parked orbit. So there was a curve, and that curve increased away from the planet, in other words, the altitude increased. I think they said it was similar to a Hohmann transfer. Yes... "Typical lunar transfer trajectories approximate Hohmann transfers... " Hohmann transfer: "The Hohmann transfer orbit is one half of an elliptic orbit that touches both the orbit that one wishes to leave (labeled 1 on diagram) and the orbit that one wishes to reach (3 on diagram). The transfer (2 on diagram) is initiated by firing the spacecraft's engine in order to accelerate it so that it will follow the elliptical orbit; this adds energy to the spacecraft's orbit. " Its that 2, or yellow part of the trajectory I'm curious about. Imagine if that RED line represents the outer edge of the Radiation Belt. See that, thats not a straight line out of the belt, thats a curved line. That means it takes longer to pass through. The incline took it into the VABs, everyone agrees about that, but the curved line keeps it there longer than if it was a straight line out! "The next step on our journey to the moon is the Trans Lunar Injection, known as the TLI for short. A single burn of the SIVB engine will allow the spacecraft to escape the EPO and extend the orbit in order to reach the moon." Lets take a look at that picture again. The one I linked several times. upload.wikimedia.org/wikipedia/en/thumb/0/03/Tli.svg/400px-Tli.svg.pngSee where that TLI happened (red dot) That happened behind the planet if the target (moon) was on the opposite side of the planet- remember, 1.5 revolutions. That means there was at least .5 of distance that the craft had to traverse, so at least a half a revolution around the earth before it started its course to the moon. But at that point the orbit started to extend, and from what I can see it only took about 5 minutes for it to go in the inner radiation belt. So again, and nobody has answered this. How long was Apollo in the VABs? I dont care if it was barely touching the edge or it didnt go through the hot spots. I asked how long it was in the VABs and where did they get this 1 hr figure from?
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Post by JayUtah on Nov 12, 2009 15:19:50 GMT -4
I'm being too demanding? Yes, to the point where your sincerity is no longer credible. You cant possibly tell me that a 2D drawing can take everything into consideration what happened in a 3D environment.That's why I created a three-dimension model, photographed it, and provided you with the photographs. Further, a series of 2D drawings accompanied by appropriate explanation suffices in every orbital mechanics class and textbook in the world. What makes you so special? What you are showing me that drawing is after TLI, which occurred halfway around the Earth- 1.5 orbits before injection- that the craft simply went straight? Because that's what your diagram suggests.No, it does not. Jason draws the plane of the transfer orbit seen edge-on, therefore as a line. All points in the orbit lie within that plane. While some of the points in that plane fall within the dangerous portion of the Van Allen belts, I showed you how an orbit can be described in that plane that doesn't require passing through them. An ellipse and a torus need not necessarily intersect if the plane of the ellipse is inclined, as Jason has illustrated. See that, thats not a straight line out of the belt, thats a curved line.The only one who has ever suggested that the Apollo spacecraft took a straight-line path was you, and you were immediately corrected. Jason knows orbital mechanics. Bob knows orbital mechanics. I know orbital mechanics. The only one who cannot demonstrate competence in orbital mechanics is you. How do you plan to mitigate that shortcoming? So again, and nobody has answered this. How long was Apollo in the VABs?You were asked what you considered an appropriate minimum threshold of exposure because the belts are not sharply defined but instead highly variable. You whine that no one has answered your question, but you ignore where people have pointed out that your question lacks enough information for a meaningful answer to be given. You don't get to pound your fist and demand an answer to an oversimplified, incomplete question. The one-hour figure derives from one person's notion of significant exposure. What is your criteria for significant exposure? I would like a number, please.
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Bob B.
Bob the Excel Guru?
Posts: 3,072
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Post by Bob B. on Nov 12, 2009 15:26:27 GMT -4
... that the craft simply went straight? Because that's what your diagram suggests. No, the drawing shows the orbit edge-on, therefore it looks like a straight line. Any change in aspect from that shown and the orbit would appear as an ellipse.
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Post by JayUtah on Nov 12, 2009 15:26:57 GMT -4
But at that point the orbit started to extend, and from what I can see it only took about 5 minutes for it to go in the inner radiation belt.That drawing does not show distances and sizes to scale. Conceptually, the lowest portion of the transfer orbit as depicted in that drawing would be where the spacecraft skirted underneath the most intense portion of the Van Allen belts. In that drawing the rough plane of the Van Allen belts would be about 11 degrees inclined to the equatorial plane, but in an orientation that's not depicted in your drawing. But this means that "underneath" can be a shallow dip if the launch and TLI are timed appropriately.
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Post by jaydeehess on Nov 12, 2009 15:45:18 GMT -4
Please illuminate a poor electronic tech. In that polar coordinate system, where is the origin considered to be? The Earth center? Reference plane, the ecliptic? .......... In this particular case, I'd place the origin at the center of the Earth. Polar coordinates were taught in the physics classes I took for the electronics diploma. I assumed that Earth's center would be you origin(as oppsoed to the Sun's center) and that one reference plane would be the ecliptic, what's the other one? It would be perpendicular to the ecliptic, so through the poles at the time of launch? the problem I am having is that this will not take into account the movement around the Sun or rotation of the Earth and that this might make the lines drawn look odd.
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Post by scooter on Nov 12, 2009 15:49:29 GMT -4
Some basics here...
You mentioned earlier about the possibility of the Apollo flights reversing course during the outbound leg, indeed it is feasible (at the proper point in the trip) for that reverse course to have been done. It was a recovery option for the Apollo 13 scenario, discarded due to the uncertain condition of the SPS.
Then you mentioned burning the engine in lunar orbit in order to come home...that is precisely how TEI worked, they burned on the far side of the Moon, and their trajectory led to Earth capture and reentry.
As for the TLI, they launched into low Earth orbit (LEO)...they call that "insertion". They orbited for "about" 1.5 orbits...around Hawaii on their second orbit, they relit the SIVB engine for the "injection"...aka TLI. This accellerated them into a highly eliptical orbit, with the high point (about opposite where the burn occurred) roughly where the Moon would be in about 3 days. (taking into account the gravity of Earth, Moon and even Sun).
After the burn finished they were still rapidly accellerating away from the surface (In a curving flightpath relative to Earth, due to our gravity well), at around 30 degrees north latitude. At some point, the ground track actually reversed and started tracking westward. How long it took to get through the northern fringes of the VAB doesn't matter much for me, as it was the fringes, and they were indeed in a "shielded" spacecraft. The radiation would have fluctuated as they went through the varying belts, but the trajectory minimized any exposure.
Finally, any diagram posted will be 2D, by definition, being seen on a flat video screen. It's really up to you to extrapolate it into 3D. And the VAB cannot be seen as an "all or nothing" mass, there are areas of less radiation or more, they fluctuate with solar activity. So where they "start" or "stop" is somewhat nebulous. What is your evidence that the VAB were an inpenetrable barrier to a successful mission?
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Post by drewid on Nov 12, 2009 16:47:45 GMT -4
Can someone get me hourly coordinates for the track? Polar would be OK, XYZ would be faster for me to work with.
Edit- Actually polar would be fine, I've just found the "Make polar coordinates work" button, and more frequent than hourly would be good for the belt transition, say every ten minutes ?
Can I get these from the voice transcripts?
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Post by nedd on Nov 12, 2009 17:37:51 GMT -4
Polar coordinates were taught in the physics classes I took for the electronics diploma. I assumed that Earth's center would be you origin(as oppsoed to the Sun's center) and that one reference plane would be the ecliptic, what's the other one? It would be perpendicular to the ecliptic, so through the poles at the time of launch? the problem I am having is that this will not take into account the movement around the Sun or rotation of the Earth and that this might make the lines drawn look odd. For Earth-centered coordinates, I think he specified the equatorial plane as one and the other as a plane containing the origin, being normal to the equatorial and parallel to a line containing the Sun and Earth at Vernal Equinox.
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Ian Pearse
Mars
Apollo (and space) enthusiast
Posts: 308
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Post by Ian Pearse on Nov 12, 2009 17:44:58 GMT -4
Jay, can I have that Van Allen belt if you're finished with it? It looks delicious. I prefer the Van Allen Belts with the chocolate sprinkles on.. ;D
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Bob B.
Bob the Excel Guru?
Posts: 3,072
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Post by Bob B. on Nov 12, 2009 18:18:45 GMT -4
Can someone get me hourly coordinates for the track? Polar would be OK, XYZ would be faster for me to work with. Edit- Actually polar would be fine, I've just found the "Make polar coordinates work" button. Not at the moment. If I can figure out what the orbital elements of the translunar orbit were, I hope to calculate the spacecraft position versus time (probably in polar coordinates). If I can find the time to work on this, I'll post the results once I have them. I make no promises as to when I'll get around to it. Polar coordinates were taught in the physics classes I took for the electronics diploma. I assumed that Earth's center would be you origin(as oppsoed to the Sun's center) and that one reference plane would be the ecliptic, what's the other one? It would be perpendicular to the ecliptic, so through the poles at the time of launch? the problem I am having is that this will not take into account the movement around the Sun or rotation of the Earth and that this might make the lines drawn look odd. For Earth-centered coordinates, I think he specified the equatorial plane as one and the other as a plane containing the origin, being normal to the equatorial and parallel to a line containing the Sun and Earth at Vernal Equinox. Nedd, you're good up until that last part. For geocentric orbits, longitude is measured in the equatorial plane with zero degrees being in the direction of the vernal equinox. Latitude is the angle above or below the equatorial plane, and is measured in the plane that is normal to the equatorial plane and contains the line connecting the center of Earth to the spacecraft (this is not a fixed plane as it moves with the spacecraft). It should also be noted that the location of the vernal equinox is constantly changing due to precession of Earth's axis. Since the coordinate system is tied to the vernal equinox, whenever orbital elements are given it is necessary to specify the time for which those elements are valid.
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Post by Jason Thompson on Nov 12, 2009 18:21:35 GMT -4
You cant possibly tell me that a 2D drawing can take everything into consideration what happened in a 3D environment. No-one is saying that. In fact we've gone to great lengths to point out the limitations of even a 3D rendering. The only reliable way to model the events properly is to use animation, since everything is in motion. I don't have the time, the software or the expertise to put together such an animation. Strangely enough, that doesn't stop me being able to visualise it. Somewhere along the line you seem unable to do so. As I explained, my 3D rendering skills are limited. However, a 2D representation is adequate for most. If you can't even grasp the basic notion that I was showing the orbital planes edge on then what's the point? I even stated it numerous times and wrote 'plane' on every line. Yes, the orbit was curved, but if you look at the orbit from edeg on it looks like a straight line. I assume you have no trouble grasping that fact in everyday situations, unless you wonder what happened to your shiny round DVDs when they're turned edge on. Why don't you get it here? You seem to be being deliberately obtuse. UNLESS the curve also happened to carry it OVER or UNDER the most intense parts of the belts. Why are you having such a hard time grasping this. I'll say it again: The belts and the transfer orbit are INCLINED with respect to each other by some 40 degrees or so. If you look at it from the top down it looks like it's spending a long time in the belt. If you look at it edge on you can see it skim the upper edge and pass OVER the rest of the belt. Well you should, because that makes a HUGE difference to the radiation exposure. They got the figure from understanding the 3 dimensional relationships between the INCLINED orbits and belts, and by measuring the radiation as they passed through. You really are the ONLY person on this thread not getting the idea that it is possible to skim over or under most of the belts, despite many efforts to enlighten you.
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Post by jaydeehess on Nov 12, 2009 19:02:46 GMT -4
For Earth-centered coordinates, I think he specified the equatorial plane as one and the other as a plane containing the origin, being normal to the equatorial and parallel to a line containing the Sun and Earth at Vernal Equinox. Nedd, you're good up until that last part. For geocentric orbits, longitude is measured in the equatorial plane with zero degrees being in the direction of the vernal equinox. Latitude is the angle above or below the equatorial plane, and is measured in the plane that is normal to the equatorial plane and contains the line connecting the center of Earth to the spacecraft [/quote] Ok , got it, though since its been 25 years since I did anything with polar coordinates I wait for the finished product. Yes, the fact that the earth revolves about its axis and orbits the Sun while a spacecraft heads for the Moon which itseld is orbiting the Earth was,,,, well,,, making my head spin trying to visualize the system. Over the course of a few days travel though this would be essentially a constant. You are speaking to the differences one would encounter over several years such as if one were trying to plot the SOHO position correct? I know that my questions are a little off the thread topic but the resident HB doesn't seem to get it at all so I have been rather ignoring him and instead trying to better visualize the actual spacecraft track myself. That he thought the simplified drawing published for the main stream media was a detailed plot of the track of the craft and that Bob's VAB sketch would be to exacting scale, is quite telling, IMO. If I were to find a simplified drawing of how a nuclear power station operates I wonder if he would take that as gospel?
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Post by nedd on Nov 12, 2009 19:24:34 GMT -4
Nedd, you're good up until that last part. For geocentric orbits, longitude is measured in the equatorial plane with zero degrees being in the direction of the vernal equinox. Latitude is the angle above or below the equatorial plane, and is measured in the plane that is normal to the equatorial plane and contains the line connecting the center of Earth to the spacecraft (this is not a fixed plane as it moves with the spacecraft). It should also be noted that the location of the vernal equinox is constantly changing due to precession of Earth's axis. Since the coordinate system is tied to the vernal equinox, whenever orbital elements are given it is necessary to specify the time for which those elements are valid. Thanks, I think between you and Google I have it, now. Knowing a little better now, it doesn't appear useful to me to include a plane defined they way I did for polar coordinates.
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