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#1
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I found a web page on Edwards Air Force Base website that describes
early captive silo test launches of the Minuteman ICBM. In the article it is claimed that, "a Minuteman passes through Mach 1.0 by the time it travels its own length above the launcher." A Minuteman I is 55.9-ft tall, A Minuteman III is 59.75-ft tall. Can anyone verify this claim? The original article is at this URL: http://www.edwards.af.mil/moments/do.../59-02-17.html -- Rusty Barton - Antelope, California | E-mail - | Visit my Minuteman ICBM website at: | http://www.geocities.com/minuteman_missile | |
#3
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If S = 0.5 a t**2,
and V = at, Then since S = 55 ft, and V = 1100 ft/sec a t**2 = 110 and at = 1100 gives t= 1/10 sec a = 11000 ft/sec/sec where g = 32.2 ft/sec/sec hence a = 340 G's ....which strikes me as implausible. That would mean it reaches its burnout velocity of approx 22,000 ft/sec in two seconds. I don't think so. "Rusty B" wrote in message om... I found a web page on Edwards Air Force Base website that describes early captive silo test launches of the Minuteman ICBM. In the article it is claimed that, "a Minuteman passes through Mach 1.0 by the time it travels its own length above the launcher." A Minuteman I is 55.9-ft tall, A Minuteman III is 59.75-ft tall. Can anyone verify this claim? The original article is at this URL: http://www.edwards.af.mil/moments/do.../59-02-17.html -- Rusty Barton - Antelope, California | E-mail - | Visit my Minuteman ICBM website at: | http://www.geocities.com/minuteman_missile | |
#4
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![]() Rand Simberg wrote: Seems dubious to me. Unless I screwed up the calculation, that means over 300 g acceleration (zero to 760 mph in sixty feet). I'm pretty sure its liftoff T/W is nowhere near that... The only one I can think of that would do that is the Sprint ABM. Pat |
#5
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On Wed, 22 Oct 2003 18:05:46 GMT, in a place far, far away, "James
Oberg" made the phosphor on my monitor glow in such a way as to indicate that: If S = 0.5 a t**2, and V = at, Then since S = 55 ft, and V = 1100 ft/sec a t**2 = 110 and at = 1100 gives t= 1/10 sec a = 11000 ft/sec/sec where g = 32.2 ft/sec/sec hence a = 340 G's ...which strikes me as implausible. That would mean it reaches its burnout velocity of approx 22,000 ft/sec in two seconds. Actually, it would do much better than that, since the 340 g is just the *beginning* acceleration, when it's fully fueled. Just imagine what it would be as it approaches burnout. ;-) -- simberg.interglobal.org * 310 372-7963 (CA) 307 739-1296 (Jackson Hole) interglobal space lines * 307 733-1715 (Fax) http://www.interglobal.org "Extraordinary launch vehicles require extraordinary markets..." Swap the first . and @ and throw out the ".trash" to email me. Here's my email address for autospammers: |
#6
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James Oberg wrote:
If S = 0.5 a t**2, and V = at, Then since S = 55 ft, Is that the Babylonian foot or the Roman foot? [ducks and runs] |
#7
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I note in passing that the previous posts in this thread are an
admirable example of the advantages of numeracy. Take a claim that has numerical implications, crank the numbers, and see if the claim is plausible. Would that people did that more often. |
#8
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(Rusty B) wrote in message . com...
I found a web page on Edwards Air Force Base website that describes early captive silo test launches of the Minuteman ICBM. In the article it is claimed that, "a Minuteman passes through Mach 1.0 by the time it travels its own length above the launcher." A Minuteman I is 55.9-ft tall, A Minuteman III is 59.75-ft tall. Can anyone verify this claim? The original article is at this URL: http://www.edwards.af.mil/moments/do.../59-02-17.html I've heard similar claims about the shuttle, although not near as much accelaration, I recall on occasion hearing 100 mph as it clears the tower does anybody an accurate figure on this? RN |
#9
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![]() "R Neutron" wrote in message om... I've heard similar claims about the shuttle, although not near as much accelaration, I recall on occasion hearing 100 mph as it clears the tower does anybody an accurate figure on this? RN That figure is accurate. -Kim- |
#10
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It's an unladen AFRICAN swallow-foot.
"Mary Pegg" wrote in message news ![]() James Oberg wrote: If S = 0.5 a t**2, and V = at, Then since S = 55 ft, Is that the Babylonian foot or the Roman foot? [ducks and runs] |
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