This is just really awesome. A star that may have been up to or over 300 solar masses. That is the double of what was thought of as the max. It is a little hard to think of that much mass in one star.
Link

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The victories from those skilled in warfare are not considered of great wisdom or courage, because their victories have no miscalculations
- SUN-TZU

wow... that is big... lets see if 1,000,000 earths can fit inside our sun then 300,000,000 would fit inside that single star...

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May you never feel a Forgotten Dragon's Ire I miei Amici.

Actually, the sun is just over 300,000 times the mass of Terra.

To put it's size into perspective, you'd have to be over 3000AU - 1AU is the distance from Terra to the sun - from to get the same level of sunlight we get here on Earth. In comparison, Neptune is 30AU away.

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Formerly DragonRider. Almost teenage me could have been more imaginative with names.

Well... i was talking size in volume not mass... so mine was just a statement of how many earths could fit inside the sun

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May you never feel a Forgotten Dragon's Ire I miei Amici.

What are you talking about? That is more towards Luminosity than mass, and this star is extreamly bright. I has a Luminosity of around (8.7) × 10^6 times the luminosity of the sun. And things like that are not linear.

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The victories from those skilled in warfare are not considered of great wisdom or courage, because their victories have no miscalculations
- SUN-TZU

No, I was talking about it's luminosity. Just saying how bright the star is.

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Formerly DragonRider. Almost teenage me could have been more imaginative with names.

But your logic is still not sound. The sun is 332900 times the mass of 'Terra'. The distance would need to be about 685 AU away to be about the same apparent magnitude as the Sun.

Starting with this refined formula: Gotten from here

mstar = − 2.72 − 2.5 · log(Lstar/diststar^2)

I put -26.73 (the apparent magnatude of the sun as we see it at 1AU) in as mstar

Then put 8700000 in for Lstar (the luminosity of R136a1)

It will give the distance (diststar) for a set star with the given luminosity and apparent magnatude.

-26.73=-2.27 -2.25*log(8700000/diststar^2)
-26.73+2.27=-2.25*log(8700000/diststar^2)
-24.46=-2.25*log(8700000/diststar^2)
-24.46/-2.25=log(8700000/diststar^2)
10.87=log(8700000/diststar^2)
10^10.87= 8700000/diststar^2
74131024130.09=8700000/diststar^2
diststar^2=8700000/74131024130.09
diststar^2= 0.0001173958
diststar= 0.0001173958^-2
diststar= 0.01083327 Light-years.
diststar=63,241.1(AU)*0.01083327 (Ly)
diststar=685 AU


(And sorry I am bad at short cutting my zeroing out of items)
I am getting that nice upper feeling that I get when doing math like that.

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The victories from those skilled in warfare are not considered of great wisdom or courage, because their victories have no miscalculations
- SUN-TZU