Here we go again! Darn, darn, darn. If I didn't have spike around to help
me out, I don't know what I'd do. Thanks, spike.
Zen Master Spike writes:
>> Jeff Davis wrote: This is based on the fact that at 70 g's, the
>> hydrostatic pressure at the bottom of that one foot of water is the same
>> at a depth of 2300 feet of seawater.
>Hmmm, I didnt grok this. Why would not the pressure under
>one foot at 70 G be the same as under 70 feet at 1G? Why do
>we need the factor of 32?
When I did my little mental calculation (sloppy thinking, rightly
characterized) and came up with 70 g's, I didn't think it was right. But I
checked it (sort of) and decided that it was right. But, as you have
noted, and proved with your elegant observation by equivalence, that
"pressure under one foot at 70 G be the same as under 70 feet at 1 G", I
was WRONGO. What happened was that I took the pressure at 2300 feet of
seawater and converted it at one atm per 33 feet to 70 atm, and then
carelessly used that number as the equivalent G force. That would have
been correct only if our mars bound astronaut had been submerged in a
column of seawater 33 feet deep, not one foot deep. The bad news is I have
to eat crow, and do my little "Confession of wrongness" dance (Oh, the
ignominy!), the good news is that our astronaut gets to increase his
maximum tolerable rate of acceleration by a factor of 33, bringing it up to
2310 G. Yes! This is what I intuited originally, before buying into my
blunder. So the trip to mars is going to be quite a bit faster. Just how
much faster you'll see in a moment.
ZM Spike also asks,
>Why do we need the factor of 32?
In d=1/2 at^2 [Actually, to avoid confusion this should be written d= (1/2)
at^2 ], a is g=32 ft/sec^2.
Also, ZMS notes:
>> I learned in school that the Earth is 98,000,000 miles from he sun, so
>Minor nit: average distance from Earth to the sun is about 93 million miles
>from the sun.
Right you are, and me, well, WRONG AGAIN. Getting to be like a broken
record. Now I see what the Mars Orbiter people were up against, kinda.
So , going back into the transit time calculation with 2310 G, and
93,000,000 miles, we get
t=time to the halfway point=
=SQRT((2 x 24,366,000 mi x 5280 ft/mi)/(2310 x 32 ft/s^2))
=31 min 6 sec
Total trip time then would be: 1 hr 2 min 12 sec.
"Due to its brevity there will be no meal service during this flight.
Please remain in your pressure vessel until we have docked at the airlock."
And, for those who might be interested in the maximum velocity achieved,
v(max) = acceleration times time to midpoint
=(2310 x 32 ft/sec^2) x (1866 sec)
= 0.14 c
Best, Jeff Davis
"It's easier with a zen master backing you up."
This archive was generated by hypermail 2b30 : Mon May 28 2001 - 09:56:17 MDT