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(Created page with "ComicBookMyths, I will have to delete to your page because it is not a proper main page. It should done your own blog, but here is to text for you to copy to your Blog. Your ...")
 
 
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In Action Comics (1938) Issue #847, "Intermezzo", Jonathan Kent is retelling Martha Kent an adventure he once had with Clark in order to reassure her that the Man of Steel will be fine in the Phantom Zone. During the flashback Jon and Clark are in the middle of space when Clark heres distress calls from members of a distant star system and races over there to help them out. While reporting on the battle, Jon says that he was light-years away and was yet still able to see and hear the battle and Superman was with him before flying over meaning he must've covered this distance to reach the aliens.
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In [[File:Sun_eater_1.jpg|thumb|left]]Action Comics (1938) Issue #847, "Intermezzo"[[File:Sun_eater_2.jpg|thumb]], Jonathan Kent is retelling Martha Kent an adventure he once had with Clark in order to reassure her that the Man of Steel will be fine in the Phantom Zone. During the flashback Jon and Clark are in the middle of space when Clark heres distress calls from members of a distant star system and races over there to help them out. While reporting on the battle, Jon says that he was light-years away and was yet still able to see and hear the battle and Superman was with him before flying over meaning he must've covered this distance to reach the aliens.
   
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<gallery widths="310" position="center" spacing="small">
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Sun eater 3.jpg
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Sun eater 4.jpg
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Sun eater 5.jpg
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sun eater 6.jpg
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sun eater 7.jpg
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</gallery>
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Now the distance claim is vague and therefore ALMOST unquantifiable, but luckily for us we can figure it out using information from what happens later on in the comic. During their journey on space they saw Kepler's supernova.
   
Now the distance claim is vague and therefore ALMOST unquantifiable, but luckily for us we can figure it out using information from what happens later on in the comic. During their journey on space they saw Kepler's supernova:
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And after the battle is over, Jon reports that the size of the explosion caused by Superman carrying a bomb into the Sun-eater was 50 times the size of Kepler's supernova:[[File:Sun_eater_8.jpg|thumb|left]]
   
And after the battle is over, Jon reports that the size of the explosion caused by Superman carrying a bomb into the Sun-eater was 50 times the size of Kepler's supernova:
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And Jon was not hit by the explosion but was rather commenting on it from further away, meaning he would've had to be further away from the battl[[File:Sun_eater_9.jpg|thumb|286px]]e scene than the radius of that explosion. Now, how big is Kepler's supernova? Well according to http://chandra.harvard.edu/photo/2012/kepler/, the image scales 5 arcminutes across, an arc minute being equal to a 60th of a degree (in terms of angle, not temperature). This would translate to a size between 19 and 33 light-years. On average this would make Kepler's supernova 26 light-years big, if this is assumed to be its diameter then the radius would be half of that, i.e. 13 light-years. Multiplying that by 50 for the radius of the explosion we get 650 light-years. Now, this still isn't the final distance as remember Jon was far away from the explosion, and since he used light-years as a whole number, it's fair to assume he was an extra light-year away from the explosion. And not to mention that afterwards the ship came flying to Clark's rescue anyway:
   
And Jon was not hit by the explosion but was rather commenting on it from further away, meaning he would've had to be further away from the battle scene than the radius of that explosion. Now, how big is Kepler's supernova? Well according to http://chandra.harvard.edu/photo/2012/kepler/, the image scales 5 arcminutes across, an arc minute being equal to a 60th of a degree (in terms of angle, not temperature). This would translate to a size between 19 and 33 light-years. On average this would make Kepler's supernova 26 light-years big, if this is assumed to be its diameter then the radius would be half of that, i.e. 13 light-years. Multiplying that by 50 for the radius of the explosion we get 650 light-years. Now, this still isn't the final distance as remember Jon was far away from the explosion, and since he used light-years as a whole number, it's fair to assume he was an extra light-year away from the explosion. And not to mention that afterwards the ship came flying to Clark's rescue anyway:
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So all in all, Clark flew 651 light-years to reach the battle scene and he did this in a time much shorter than 651 years, in other words this is a definite FTL feat, no denying it. And on top of that, if you think physics and calculations shouldn't be applied to comics, then think again cos writer Dwayne McDuffie had a Masters degree in applied physics hence all the science references: https://io9.gizmodo.com/the-wife-of-legendary-comics-writer-dwayne-mcduffie-wan-1789270102.
 
So all in all, Clark flew 651 light-years to reach the battle scene and he did this in a time much shorter than 651 years, in other words this is a definite FTL feat, no denying it. And on top of that, if you think physics and calculations shouldn't be applied to comics, then think again cos writer Dwayne McDuffie had a Masters degree in applied physics hence all the science references: https://io9.gizmodo.com/the-wife-of-legendary-comics-writer-dwayne-mcduffie-wan-1789270102.
 
   
 
Now to figure out the time, we know it definitely didn't take him centuries to reach there because Jon would have been dead and so would the aliens under attack but we do have one reference to time from Clark. In the first page from the comic that I've shown, as he's heading off he tells the ship to take Jon home if he's not back within 3 hours. Here's the scan again:This is neat but it doesn't seem like an accurate reference of time because the ship doesn't actually take Jon home but rather goes towards Clark meaning 3 hours hadn't passed. Not to mention that the aliens under attack would most likely not be able to hold off the Sun-eater for 3 hours, this was most likely a nod to how long Clark thought the actual battle might take and not how long he would take to reach there. However I will be using this time frame as a low end to calculate the feat, despite how unlikely it is. So if Superman really did take 3 hours to cover 651 light-years then his average speed would still be:
 
Now to figure out the time, we know it definitely didn't take him centuries to reach there because Jon would have been dead and so would the aliens under attack but we do have one reference to time from Clark. In the first page from the comic that I've shown, as he's heading off he tells the ship to take Jon home if he's not back within 3 hours. Here's the scan again:This is neat but it doesn't seem like an accurate reference of time because the ship doesn't actually take Jon home but rather goes towards Clark meaning 3 hours hadn't passed. Not to mention that the aliens under attack would most likely not be able to hold off the Sun-eater for 3 hours, this was most likely a nod to how long Clark thought the actual battle might take and not how long he would take to reach there. However I will be using this time frame as a low end to calculate the feat, despite how unlikely it is. So if Superman really did take 3 hours to cover 651 light-years then his average speed would still be:
   
'''651 years/3 hours = 1,902,181.37×'''''' FTL (~1.9 million) - MFTL+'''
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'''651 years/3 hours = 1,902,181.37×'''''' FTL (~1.9 million) - MFTL+'''
   
What I think was more likely was that this all happened in real time, if that was the case then the time between Superman starting to fly and arriving on the battle scene was around 19.68s. This shorter timeframe makes a lot more sense to me along with the given context. If Superman covered 651 light-years in 19.68 seconds then his average speed would be: 
+
What I think was more likely was that this all happened in real time, if that was the case then the time between Superman starting to fly and arriving on the battle scene was around 19.68s. This shorter timeframe makes a lot more sense to me along with the given context. If Superman covered 651 light-years in 19.68 seconds then his average speed would be:
   
'''651 years/19.68s = 1,043,880,020× FTL (~1 billion) - MFTL+'''
+
'''651 years/19.68s = 1,043,880,020× FTL (~1 billion) - MFTL+'''
   
 
In short, whichever way you look at it, this is a relatively casual massively faster-than-light speed for Superman which no-one cn deny. Most people took little notice of this feat and didn't bother looking into it too much, I haven't seen any other calculation made for this feat and therefore am lead to believe I am the first to do so. I would appreciate if you guys could spread this calculation around so that it could get evaluated by a member of the Calc Group and get people talking about this missed feat. Thank you.<ac_metadata title="About your calc page"> </ac_metadata>
 
In short, whichever way you look at it, this is a relatively casual massively faster-than-light speed for Superman which no-one cn deny. Most people took little notice of this feat and didn't bother looking into it too much, I haven't seen any other calculation made for this feat and therefore am lead to believe I am the first to do so. I would appreciate if you guys could spread this calculation around so that it could get evaluated by a member of the Calc Group and get people talking about this missed feat. Thank you.<ac_metadata title="About your calc page"> </ac_metadata>

Latest revision as of 00:11, December 26, 2018

ComicBookMyths, I will have to delete to your page because it is not a proper main page. It should done your own blog, but here is to text for you to copy to your Blog. Your Blogs on profile and it is the tab near your message wall. I hope this help you.


In
Sun eater 1
Action Comics (1938) Issue #847, "Intermezzo"
Sun eater 2
, Jonathan Kent is retelling Martha Kent an adventure he once had with Clark in order to reassure her that the Man of Steel will be fine in the Phantom Zone. During the flashback Jon and Clark are in the middle of space when Clark heres distress calls from members of a distant star system and races over there to help them out. While reporting on the battle, Jon says that he was light-years away and was yet still able to see and hear the battle and Superman was with him before flying over meaning he must've covered this distance to reach the aliens.

Now the distance claim is vague and therefore ALMOST unquantifiable, but luckily for us we can figure it out using information from what happens later on in the comic. During their journey on space they saw Kepler's supernova.

And after the battle is over, Jon reports that the size of the explosion caused by Superman carrying a bomb into the Sun-eater was 50 times the size of Kepler's supernova:
Sun eater 8
And Jon was not hit by the explosion but was rather commenting on it from further away, meaning he would've had to be further away from the battl
Sun eater 9
e scene than the radius of that explosion. Now, how big is Kepler's supernova? Well according to http://chandra.harvard.edu/photo/2012/kepler/, the image scales 5 arcminutes across, an arc minute being equal to a 60th of a degree (in terms of angle, not temperature). This would translate to a size between 19 and 33 light-years. On average this would make Kepler's supernova 26 light-years big, if this is assumed to be its diameter then the radius would be half of that, i.e. 13 light-years. Multiplying that by 50 for the radius of the explosion we get 650 light-years. Now, this still isn't the final distance as remember Jon was far away from the explosion, and since he used light-years as a whole number, it's fair to assume he was an extra light-year away from the explosion. And not to mention that afterwards the ship came flying to Clark's rescue anyway:

So all in all, Clark flew 651 light-years to reach the battle scene and he did this in a time much shorter than 651 years, in other words this is a definite FTL feat, no denying it. And on top of that, if you think physics and calculations shouldn't be applied to comics, then think again cos writer Dwayne McDuffie had a Masters degree in applied physics hence all the science references: https://io9.gizmodo.com/the-wife-of-legendary-comics-writer-dwayne-mcduffie-wan-1789270102.

Now to figure out the time, we know it definitely didn't take him centuries to reach there because Jon would have been dead and so would the aliens under attack but we do have one reference to time from Clark. In the first page from the comic that I've shown, as he's heading off he tells the ship to take Jon home if he's not back within 3 hours. Here's the scan again:This is neat but it doesn't seem like an accurate reference of time because the ship doesn't actually take Jon home but rather goes towards Clark meaning 3 hours hadn't passed. Not to mention that the aliens under attack would most likely not be able to hold off the Sun-eater for 3 hours, this was most likely a nod to how long Clark thought the actual battle might take and not how long he would take to reach there. However I will be using this time frame as a low end to calculate the feat, despite how unlikely it is. So if Superman really did take 3 hours to cover 651 light-years then his average speed would still be:

'651 years/3 hours = 1,902,181.37×' FTL (~1.9 million) - MFTL+

What I think was more likely was that this all happened in real time, if that was the case then the time between Superman starting to fly and arriving on the battle scene was around 19.68s. This shorter timeframe makes a lot more sense to me along with the given context. If Superman covered 651 light-years in 19.68 seconds then his average speed would be:

651 years/19.68s = 1,043,880,020× FTL (~1 billion) - MFTL+

In short, whichever way you look at it, this is a relatively casual massively faster-than-light speed for Superman which no-one cn deny. Most people took little notice of this feat and didn't bother looking into it too much, I haven't seen any other calculation made for this feat and therefore am lead to believe I am the first to do so. I would appreciate if you guys could spread this calculation around so that it could get evaluated by a member of the Calc Group and get people talking about this missed feat. Thank you.

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