Post by aramis on Jul 29, 2013 20:13:52 GMT -5
I got to thinking, "How big should an F&E hex really be?"
And that lead me to thinking, "How big was the Enterprise's 5YM?"
And that lead me to thinking about the SFTM map and it's error.
If we assume that
And knowing that circumference =2*Pi*R
we can work out the radius of the federation.
WF6 = 216C
5Y= 1080LY
distance = 2+(10/8 * Pi) = 2+3.92 = 5.92 radii
radius = 182LY
There are several other things we can do that adjust the figures...
count the inbound/outbound for TiY and A:E as a single or as two trips, figure high speed "distance routes" are at WF+1 or WF+2, and figure that the SFTM got the cruise wrong, and it was actually 6.5 or 7. Or we can figure with Speed(C)=WF^(10/3) as for the lower end of the TNG/DS9/Voyager scale.
Fast means using WF^(10/3)
Inbound Trips are the two episodes on earth.
And, for hex sizes, same categories and calculations
[/td][/tr][/table]
There are some issues, tho. If, perhaps, it was out to Klingon space, then over to Tholian, then to Romulan, back in, and then to gorn, back in, and over to Kzinti space, it may be out, two trips, and back in, but be only 1/4 or so of the rim of the federation, instead of just over half.
Fed Radius, short rim time:
[/td][/tr][/table]
Fed Space Hex, short rim time.
[/td][/tr][/table]
And that lead me to thinking, "How big was the Enterprise's 5YM?"
And that lead me to thinking about the SFTM map and it's error.
If we assume that
- WF 6 is cruise speed (SFTM)
- Speed in C= WF^3
- The 5YM went 5/8 of a circle in 5 years
- that the majority of the 5YM time was spent on the edge
- that we need to account only for the inbound and outbound legs (ignoring the two trips to Earth: Tomorrow is Yesterday and Assignment Earth)
- That non-cruise time is made up for by travel above cruise speed
And knowing that circumference =2*Pi*R
we can work out the radius of the federation.
WF6 = 216C
5Y= 1080LY
distance = 2+(10/8 * Pi) = 2+3.92 = 5.92 radii
radius = 182LY
There are several other things we can do that adjust the figures...
count the inbound/outbound for TiY and A:E as a single or as two trips, figure high speed "distance routes" are at WF+1 or WF+2, and figure that the SFTM got the cruise wrong, and it was actually 6.5 or 7. Or we can figure with Speed(C)=WF^(10/3) as for the lower end of the TNG/DS9/Voyager scale.
WF | Straight | 1 Inbound | 2 inbound | Fast Straight | Fast 1 Inb | Fast 2 trips |
6 | 182.22 | 136.24 | 108.79 | 331.11 | 247.57 | 197.69 |
6.5 | 231.67 | 173.22 | 138.32 | 432.36 | 323.28 | 258.14 |
7 | 289.35 | 216.35 | 172.76 | 553.51 | 413.86 | 330.48 |
Fast means using WF^(10/3)
Inbound Trips are the two episodes on earth.
And, for hex sizes, same categories and calculations
[tr][td]WF | Straight | 1 Inbound | 2 inbound | Fast Straight | Fast 1 Inb | Fast 2 trips |
6 | 19.18 | 14.34 | 11.45 | 34.85 | 26.06 | 20.81 |
6.5 | 24.39 | 18.23 | 14.56 | 45.51 | 34.03 | 27.17 |
7 | 30.46 | 22.77 | 18.19 | 58.26 | 43.56 | 34.79 |
There are some issues, tho. If, perhaps, it was out to Klingon space, then over to Tholian, then to Romulan, back in, and then to gorn, back in, and over to Kzinti space, it may be out, two trips, and back in, but be only 1/4 or so of the rim of the federation, instead of just over half.
Fed Radius, short rim time:
[tr][td]WF | Straight | 1 Inbound | 2 inbound | Fast Straight | Fast 1 Inb | Fast 2 trips |
6 | 302.45 | 193.87 | 142.65 | 549.59 | 352.28 | 259.22 |
6.5 | 384.54 | 246.49 | 181.37 | 717.66 | 460.01 | 338.49 |
7 | 480.29 | 307.86 | 226.53 | 918.75 | 588.91 | 433.33 |
Fed Space Hex, short rim time.
[tr][td]WF | Straight | 1 Inbound | 2 inbound | Fast Straight | Fast 1 Inb | Fast 2 trips |
6 | 31.84 | 20.41 | 15.02 | 57.85 | 37.08 | 27.29 |
6.5 | 40.48 | 25.95 | 19.09 | 75.54 | 48.42 | 35.63 |
7 | 50.56 | 32.41 | 23.85 | 96.71 | 61.99 | 45.61 |