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Smallville Revisions

According to the Smallville Wiki this is the IRL location:

  • The Kent Farm is actually 843 248th Street, Langley, BC, Canada (Corner of 8th Ave and 248th Street). Note the farm has two entrances; the farm's "television entrance" is on 8th Ave.[37] Not even a few miles south of the property is the International Boundary shared between Canada and the United States of America. Coordinates: N 49.01622┬░, W 122.53702┬░ The farm was for sale in summer 2010.
 
Lois said it was going to rain, so they're some type of Nimbus cloud.
 
2000 m to 4000 m the. Average of 3000 m.

pi*192.31^2*3000 = 3.48557846035e8 m^3

3.48557846035e8*1.003 = 3.49603519573105e8 kg

192.31/3 = 64.10 m/s

1/12*3.49603519573105e8*64.10^2 = 1.1970453643809746292e11 Joules, City Block level

Disappointing.

Although can someone try to find the distance between the house and the lake that was linked above?
 
The lake is Buntzen Lake in British Columbia. Which according to Google Maps is like 60 kilometers away.
 
Yeah. Though you'd need to use direct distance rather than road distance since clouds don't need to follow terrain.
 
Have any idea what the results would be in that case? Spino originally got 192.31 m as a distance for length based on the hourse.

Also, since a timeframe is important, do you know how long it took the clouds over Buntzen Lake to be dispersed after Clark started blowing them away from Kent farm?
 
Probably best to leave it to Spino. We also still need a timeframe for the feat if you have any idea on what that would be.
 
Oh that's easy. Just go to the gif directly and pause it when it starts and use the foreward/back option on PC. Gives the exact start and end time. If you can't d9 it I'll do it when I get home at around 6.
 
So all you need to do is time how fast the clouds moved on screen arounf the kent farm and then simply divide that by the distance to Buntzen Lake, correct?
 
  • End = 10.85
  • Time = 3.44 seconds
So

  • ¤Ç * 40,000^2 * 3000 = 1.5079645e+13 m^3
  • 1.5079645e+13 * 1.003 = 15,124,883,935,000ÔǼ kg
  • 1/12 * 15,124,883,935,000ÔǼ * (40.000/3.44)^2
  • 1,260,406,994,583.3334 * 11,627.9069^2
  • 1.7041738e+20 Joules or 6-C
 
Was the lake being referred to in the show actually Buntzen Lake, or was that just where they filmed it?
 
Filmed it. The lake used in the show is crater lake in the canon map I linked abovem
 
Probably not to scale. Just for general location thinking about it. But I don't know of another map in the series.
 
Can't seem to find any either, the one you posted might be the best we've got.

Is smallville ever given a specified size at any point, or any wide angle shots of the town itself?
 
Well, there are maps of the place but there's never a proper shot of them.

For Smallville none of the shots give an image of the farm and crater lake. We just know the place has 45,000+ people living in the area.
 
Hm, do you think the map you posted is accurate enough for scaling, or would inflate/deflate things too much?
 
And the distance between the Kents farm and Crater Lake or the Kents farm and the rest of Smallville is never stated as I understand it?
 
Hm, bit of a conundrum then. Best we can do is calc the feat with the map you've provided and decide if its reliable to use, or just stick with the vaporisation feat.
 
I have to unsubscribe from this thread due to time constraints. You can notify me later via my message wall if you need my help after you have reached a conclusion.
 
So how should we finish this up, calcing the cloud dispersal scene with the map Qaws provided, or just sticking with the lake vaporisation feat and changing all the 6-A's to Low 7-C+?
 
Either we stick with Low 7-C+ or use the IRL distance between the sites since the map is to strange to use for calcs.
 
Would the map inflate the results? Cause if all the buildings are oversized it would lowball the feat if anything since the distances between places are being portrayed as smaller than they're actually supposed to be.
 
So going by a rough pixel calc with the map

  • ¤Ç * 1,088.84125ÔǼ^2 * 3000 = 11,173,783,653.8669 m^3
  • 11,173,783,653.8669 * 1.003 = 11,207,305,004.8285 kg
  • 1/12 * 11,207,305,004.8285ÔǼ * (1,088.84125/3.44)^2
  • 933,942,083.7357 * 316.5236^2
  • 93,569,032,391,662.3681 Joules or 7-C
 
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