Text 28617, 89 rader
Skriven 2007-05-05 18:59:58 av John Hull (1:123/789.0)
     Kommentar till en text av JIM HOLSONBACK (1:123/140)
Ärende: Muslems
===============
JIM HOLSONBACK -> JEFF BINKLEY wrote:
 JH> Hi, Jeff.

 -=>> On 05-05-07 14:37 JEFF BINKLEY wrote to VERN HUMPHREY <=-

 JB>>> Actually it isn't as difficult as you are making it.  If you
 JB>>> assume that your braking deceleration is the same as your
 JB>>> acceleration then you simply divide the distance by half to
 JB>>> calculate the midway point of acceleration.  Once you have the
 JB>>> answer then you multiply by 2 to get the total time.  Mass and
 JB>>> thrust are irrelevant because Vern said acceleration was constant.
 JB>>> acceleration Of course where does the force of 1G come from ?
 JB>>> acceleration Well Vern will have to let us know that trick.

 JH> Magic thrusters!

 VH>> An acceleration of 1G is crucial because the passengers are humans.
 VH>> If it were much less than 1G, they would lose bone mass on the long
 VH>> trip. If it were much greater than 1G, they would have heart and
 VH>> circulatory problems.

 JH> Exercise to relieve boredom, prevent muscle atrophy and stress the
 JH> skeletal system. Don't astronauts do that now on the space station?

A ship built with a rotating ring around a central axis fuselage would provide
sufficient gravity to equal 1 g.  That would provide the rotational element our
bodies are tuned for, and the thrust would provide the acceleration necessary
for normal movement. Normal activity would maintain body health.

 VH>> Remember, to reach the speed of light at 1G acceleration will take a
 VH>> bit more than 58 years.  That's a long time to pack an extra 50 or
 VH>> 100 lbs of body weight.

 JH> I dunno where Vern gets this 58 years stuff, but it doesn't calculate at
 JH> that.  And what extra body weight?  While accelerating/decelerating at
 JH> 1g our bodies would seem to weigh the same as they do here on earth,
 JH> since f=ma.

 JB>> There's something wrong with either your math or mine.  I calculated
 JB>> 347 days at 1G acceleration.  I went on 10 m/s^2 acceleration and
 JH> light
 JB>> speed being 300,000 km/s.  Thus it takes 30 million seconds or 347
 JB>> days.

 JH> That was correct -

 JB>> Ok, so I went back and looked and I am off by a factor of 100.  30
 JB>> million seconds is 34722 days or 95.2 years .  Not sure what
 JH> happened
 JB>> the first time.

 JH> Nope.
 JH> 60 sec/min x 60 min/hr x 24 hr/day x 347 days =  29,980,800, say 30 mln
 JH> seconds.  You were right the first time.

 JH> Now since 347 days = 0.95 years, and with using v=at the increase in
 JH> velocity is linear over time, with a Vo=0 and V1 =  speed of light, the
 JH> average velocity over those 347 days is half the speed of light. The
 JH> distance covered while accelerating was 0.5 c x 0.95 year = 0.475 light
 JH> years.

 JH> Using the same 1g for deceleration would take another 0.95 years, and
 JH> cover another 0.475 light years. So with 2 x 0.475 light years = 0.95
 JH> light years covered starting and stopping, 9.05 light years remain to be
 JH> covered.

 JH> If it were possible that our hypothetical spacecraft could travel at
 JH> the speed of light, total time to travel 10 light years would be -

 JH>    0.95 years to accelerate at 1g from v=0 to speed of light
 JH>    9.05 years at speed of light
 JH>  + 0.95 years to declerate at 1g from speed of light to v=0
 JH>  ===========
 JH>   10.95 years total travel time to cover 10 light years

You wouldn't be traveling in the middle at light speed.  You are traveling at
.5 C after approximately 1 year.  You will continue to accelerate at the rate
of 1 g.  After covering another 4 lightyears, approximately, you would flip
ends and begin deceleration.  As I said, the two websites I found both agreed
on roughly fifteen years to travel to and return to the nearest star, which is
somewhat less than 10 light years.

-- 
John
..
My Blog: Mad Gorilla's Jungle:      http://madgorilla.terapad.com/

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