Bob B.
Bob the Excel Guru?
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Post by Bob B. on Jul 10, 2007 15:29:48 GMT -4
Pete, just to expand a bit on one of Bob's points - one reason this is done is to control the aerodynamic loading on the vehicle. You'll hear the PAO (voice commentator) mentioning the SSMEs throttling down as "Max Q" is approached. Yeah.. I was going to mention this but I had to run and was forced to cut my post short. The SSME (Space Shuttle Main Engines) start at 104% of rated thrust at liftoff but are throttled back to 65% at about T+35 seconds as the aerodynamic loads on the vehicle begin to build toward maximum Q, the momentum of greatest aerodynamic load. Max Q occurs at around T+60 seconds and then the SSME are throttled back up to 104% at about T+65 seconds as the as loads subside. In regard to my second point, any engine produces more thrust in a vacuum than it does when there is an ambient air pressure. When in the atmosphere, the vacuum thrust of an engine is decreased by the air pressure times the area of the nozzle exit. The engines used on a booster generally have two thrust ratings, one at sea level and the other in a vacuum. As I recall, the SSME have a rating of 375,000 lbf each at sea level and 470,000 lbf each in a vacuum. As the Shuttle ascends through the atmosphere, the thrust progressively increases. (As a side note, since we know atmospheric pressure at sea level is 14.7 PSI, we can easily calculate the diameter of the SSME nozzle from the difference in the two thrust ratings. Anyone care to give it a try?) My third point addressed the thrust of the SRBs (Solid Rocket Boosters). The gas production rate and chamber pressure of the SRBs, and hence the thrust, is dependent on the area of the burn surface of the propellant block. As propellant burns away, the contours of this surface are constantly changing. Burning usually occurs along the surface of a central channel running down the length of the propellant block. The shape of this channel can control how the burn progresses, and thus how the thrust varies. The channel shape can be designed to give high thrust at the beginning, high thrust at the end, or relatively constant thrust. The Shuttle SRBs use a combination of shapes that yields maximum thrust shortly after ignition, which then gradually steps down as the burn continues. There is then a transient period of several seconds at the end when the thrust drops off to zero.
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Post by ishmael on Jul 10, 2007 15:35:41 GMT -4
The SSME (Space Shuttle Main Engines) start at 104% of rated thrust at liftoff but are throttled back to 65% at about T+35 seconds as the aerodynamic loads on the vehicle begin to build toward maximum Q, the momentum of greatest aerodynamic load. Max Q occurs at around T+60 seconds and then the SSME are throttled back up to 104% at about T+65 seconds as the as loads subside. Not my line of work. But, the way I would think of it is, moving faster means more friction. But moving higher means thinner air, so less friction. So the maximum load would occur when the effect from increasing speed is exactly offset by the effect from thinning air. Is that the right way to think about it?
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Post by petereldergill on Jul 10, 2007 15:37:18 GMT -4
Yeah.. I was going to mention this but I had to run and was forced to cut my post short. . I like your first one better, it was easier for me to understand ;D And all this time I thought rocketry was easyPete
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Post by gwiz on Jul 10, 2007 16:23:30 GMT -4
Not my line of work. But, the way I would think of it is, moving faster means more friction. But moving higher means thinner air, so less friction. So the maximum load would occur when the effect from increasing speed is exactly offset by the effect from thinning air. Is that the right way to think about it? Pretty much so. The "q" in "max q" is dynamic pressure, which varies as air density times the square of velocity. Other things being equal, aerodynamic loads are proportional to q. Throttling down the Shuttle engines reduces velocity for a given height, hence air density, and thus gives a lower value for max q.
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Bob B.
Bob the Excel Guru?
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Post by Bob B. on Jul 10, 2007 16:57:02 GMT -4
Not my line of work. But, the way I would think of it is, moving faster means more friction. But moving higher means thinner air, so less friction. So the maximum load would occur when the effect from increasing speed is exactly offset by the effect from thinning air. Is that the right way to think about it? Yep, that's essentially correct. Mathematically the drag force is given by the equation, F D = C D rho V 2 A / 2 where C D is drag coefficient, rho is air density, V is velocity, and A is area normal to the flow. Air density decreases with altitude but the rocket goes faster the higher it travels. Therefore as a function of time, rho decreases and V increases. Also C D is not constant. Drag coefficient varies as a function of the Mach number and peaks at about Mach 1.5. Initially C DV 2 increases faster than rho decreases, thus the drag force increases. However, after max Q the drag coefficient begins to decrease and C Drho decreases faster than V 2 increases, thus drag force decreases. Eventually C D levels off but rho continues to decrease to practically nothing. By the time the rocket reaches orbit the air density is extremely low, but there is still a small drag force because of the very high velocity. The drag is enough that over time the orbit decays.
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Post by JayUtah on Jul 10, 2007 17:19:08 GMT -4
And all this time I thought rocketry was easy
Easy rocketry is easy in the same way that easy cooking is easy. The basics are simple enough to understand, but when it's necessary to carry it to a high art, additional complexity and refinement are needed.
Where margins are small and accuracy counts, many second- and third-order effects must be considered. For basic rocketry you can worry only about first-order effects and allow yourself wide margins.
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Post by Count Zero on Jul 10, 2007 21:05:33 GMT -4
In high school physics it is common to work problems dealing with ballistic trajectories using d = vt + .5 at^2. It is usually assumed that the ground is a flat surface and gravity is always acting straight downward at constant magnitude. In this simplified case the trajectory is a parabola. Now visualize that the surface is curved and gravity is directed radially inward toward the center of curvature and weaker with increasing distance. Your perfect parabolic trajectory is now an ellipse. - Bob B.
That is a really excellent explanation. Thanks!
(edited to attribute the quote)
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Post by Mr Gorsky on Jul 11, 2007 5:04:40 GMT -4
And suddenly this discussion makes sense. To whoever provided the explanation of the effects of gravity on a ballistic trajectory (as quoted by Count Zero) my thanks.
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Post by BertL on Jul 12, 2007 17:19:58 GMT -4
I've done some more analysing on a screenshot in this picture. Peculiarly enough the debator keeps insisting that the discolorisation that is completely 100% consistant with compression artifacts is a wire, for some reason. I mean, come on, it's obvious that the discolorisation is from compression, but for some reason this specific discolorisation is not from compression, but from a wire. Seriously. You can find the discussion here, by the way. I'm starting to dislike how crappy it is to debate over YouTube.
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Post by PhantomWolf on Jul 12, 2007 18:07:59 GMT -4
Well since if clearly continues down into the PLSS backpack, yes it is a compression artifact. Ask him if he can find it on the high quality DVD versions of the footage rather then the compressed to heck and back YouTube version.
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Post by Mr Gorsky on Jul 13, 2007 5:25:34 GMT -4
Well since if clearly continues down into the PLSS backpack, yes it is a compression artifact. Ask him if he can find it on the high quality DVD versions of the footage rather then the compressed to heck and back YouTube version. Welcome to the 21st Century mindset ... anything important is on the web, so you don't need to look anywhere else. Unfortunately, being a musician, I am one of those dinosaurs who still believes that a CD is better quality than an mp3 ... and a DVD is better quality than a FLV file ... and an original, film negative is better than a JPEG. But, what do I know? I also recall another discussion on another board, when I was asked to cite my source when I quoting from an interview, and gave magazine title, date and page number only to be told that it didn't count because it couldn't be linked to and verified by anyone else on the board!
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Post by BertL on Jul 13, 2007 5:35:58 GMT -4
Hahahaha, the conspiracy theorist just admitted himself that he can't find the wire in the original frame, but he can find it in the extra-compressed YouTube version. It's still a wire though, according to him. That just makes me laugh.
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MarkS
Earth
Why is it so?
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Post by MarkS on Jul 13, 2007 8:18:30 GMT -4
As I recall, the SSME have a rating of 375,000 lbf each at sea level and 470,000 lbf each in a vacuum. As the Shuttle ascends through the atmosphere, the thrust progressively increases... (As a side note, since we know atmospheric pressure at sea level is 14.7 PSI, we can easily calculate the diameter of the SSME nozzle from the difference in the two thrust ratings. Anyone care to give it a try?) My pleasure! Delta thrust = vacuum thrust - sea level thrust = 470,000 lbs - 375,000 lbs = 95,000 lbs Nozzle area = delta thrust / air pressure at sea level = 95,000 lbs / 14.7 lbs/sq. in. = 6460 sq. in. (assuming 3 sig. digits) Nozzle diameter = SQRT(nozzle area/pi) = SQRT(6460 sq. in. / 3.14) = SQRT(2060 sq. in.) = 45.4 in. radius, or 90.8 in. diameter. Notes: - Yes, that's a FORTRAN function for the square root.
- Web search confirms SSME exit diameter of 90.7 in.
- This, plus two Charitable Acts, should hopefully earn me my Junior Shill badge.
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Bob B.
Bob the Excel Guru?
Posts: 3,072
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Post by Bob B. on Jul 13, 2007 8:27:44 GMT -4
My pleasure! Delta thrust = vacuum thrust - sea level thrust = 470,000 lbs - 375,000 lbs = 95,000 lbs Nozzle area = delta thrust / air pressure at sea level = 95,000 lbs / 14.7 lbs/sq. in. = 6460 sq. in. (assuming 3 sig. digits) Nozzle diameter = SQRT(nozzle area/pi) = SQRT(6460 sq. in. / 3.14) = SQRT(2060 sq. in.) = 45.4 in. radius, or 90.8 in. diameter. Notes: - Yes, that's a FORTRAN function for the square root.
- Web search confirms SSME exit diameter of 90.7 in.
- This, plus two Charitable Acts, should hopefully earn me my Junior Shill badge.
Well done! You're well on your way to becoming a government shill.
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Post by JayUtah on Jul 13, 2007 11:37:03 GMT -4
Welcome to the 21st Century mindset ... anything important is on the web, so you don't need to look anywhere else.
I very much enjoy instant research, but I still recognize the value of doing it the old-fashioned way.
Unfortunately, being a musician, I am one of those dinosaurs who still believes that a CD is better quality than an mp3...
I remember the same argument over vinyl versus CD. And it was defensible: CDs sounded tinny in the mid-1980s. I still prefer being in the same room as the actual performers. I remember standing on stage between the brass section and the percussion section of the Utah Symphony during a rendition of Copland's "Fanfare for the Common Man." That's quite an experience. Little beats the Edgar Thompson Recital Hall up at the University of Utah, not only because it's superbly constructed for the performance of chamber music but because Edgar Thompson is an ex-Apollo engineer. How he got a recital hall named after him is an amusing story.
...it didn't count because it couldn't be linked to and verified by anyone else on the board!
Most ironic; academic standards of web reference require not only the URL but also the date on which the URL was accessed, according to the presumption that the contents at the other end of the URL can disappear or change without warning. Whereas, e.g., Mauldin's Prospects for Interstellar Travel has remained entirely unchanged since it was first committed to squished-flat trees some 20 years ago. It simply requires getting off one's hindquarters to verify.
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