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Re: (TFT) Replies to recent topics

Sorry Thorn. I mentioned Stan in that last reply. It was your question I was sneaking out of. Here is a better answere. (I needed some time to do the math). 



David Michael Grouchy II wrote:
>
>     The answer is T=1 and G=2.



From: Thorn <edt@dopey.ne.mediaone.net>

Ah, no.  T is the number of times the object has to traverse the
distance between the gates.  Since the object has a 1/216 chance of
destabilizing the gate pair as it passes, one pair will not (likely)
provide enough transits to accelerate the object to 11Km/s.


Thorn,
  The answer is T = 307,438 and G = 2.
  Now showing my work.

Fall for 1120.2 turns
9.8 m per second squared
9.8m * 5601 * 5601
9.8m * 31,371,201 = 307,437,769.8 meters

In 1121 turns a frictionless demon will have traveled about 307,437 km
On the 5600 second of falling the demon will be going 11199 meters a second.
On the 5601 second of falling the demon will be going 11.201m per second, or escape velocity.
For the medievalist out there, this is about 25,000 miles a second. Now back to metric.
The whole trip requires 307,437.7698 km of distance to fall. Now the Earth itself is only 12,756km across. Thus the trip would encompass a distance 24 times the diameter of the Earth. 
Using gate pairs 1km apart, T= 307,438
The gates would start flickering early in the trip and go out before escape velocity could be reached. If two teams of wizards were on site, one up in the air on magic carpets, to control the gates, it might succeed. But using the figure you gave of failure 1/216 times, it is still very unlikely. They would have 1423 failures, and at a cost of 10 per control gate spell the two wizards would need 284 apprentices each who supply 25 stength. That?s a lot of flying carpets. 

May I suggest a Gauss gun laid up against the Andes Mountains instead?
The demon could also teleport into orbit.

    David Michael Grouchy II.
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