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Question for for more versed in this than I.

http://mikex.me.uk/misc/rotj/pages/rotj_mq_261.htm

How big (approximately) would the DS2 look in the Endor sky if it were 160km in diameter, and was 33,624.7 km from the surface of Endor?

Just an innocent question....

Lord Poe wrote:Question for for more versed in this than I.

http://mikex.me.uk/misc/rotj/pages/rotj_mq_261.htm

How big (approximately) would the DS2 look in the Endor sky if it were 160km in diameter, and was 33,624.7 km from the surface of Endor?

Just an innocent question....

Why 33,624.7 km?

Lord Poe wrote:Question for for more versed in this than I.

http://mikex.me.uk/misc/rotj/pages/rotj_mq_261.htm

How big (approximately) would the DS2 look in the Endor sky if it were 160km in diameter, and was 33,624.7 km from the surface of Endor?

Just an innocent question....

You took my advice on Sarli! Smooth work.

Gustav32Vasa wrote:

Lord Poe wrote:Question for for more versed in this than I.

http://mikex.me.uk/misc/rotj/pages/rotj_mq_261.htm

How big (approximately) would the DS2 look in the Endor sky if it were 160km in diameter, and was 33,624.7 km from the surface of Endor?

Just an innocent question....

Why 33,624.7 km?

This moron named M Sarli who is a SW RPG nerd and uses clown math to "refute" Dr. Saxton claims that the ROTJ film supports a DS2 which is 160 km, but nearly 34 thousand kilometers from the moon.

Edit: It would hardly be visible.

Illuminatus Primus wrote:You took my advice on Sarli! Smooth work.

Yup! Gotta take advantage of the visual medium!

An object distance D away (observer to center of object) and d diameter takes up, when D >> d, about 2*atan(d/2D) radians of the field of view. Really, d/D is plenty good enough.

In this case, the Death Star II takes up 160/33624.7 = a bit under .0048 radians of the sky, or over 0° 16'. For comparison, this is about half the size of the Moon in our sky.

EDIT: The exact formula is 2*asin(d/2D), BTW.

Darth Holbytlan wrote:In this case, the Death Star II takes up 160/33624.7 = a bit under .0048 radians of the sky, or over 0° 16'. For comparison, this is about half the size of the Moon in our sky.

That would be 1,738km in diameter? (for our moon) How visible would the DS2 (at 160 km diameter) be from Earth at that distance?

Well Luna is 1737.4 km in radius which is 3474.8 km in diameter.

Luna is 384,400 km(mean distance) from the center of the Earth.

2*atan(d/2D) = (man I can't find my calculator! The windows calculator is useless Someone please finish!)

DSII is 160 km at the same distance it would be....

Lord Poe wrote:That would be 1,738km in diameter? (for our moon)

No, the Moon is 3476 km in diameter. Or do you mean that it would appear to be roughly 1,700 km in diameter if it were the Moon's distance away? The latter is correct, although usually sky observations are measured as a viewing angle in degrees, minutes, and seconds of arc, since that's distance independent. (A second of arc is 1/60th of a minute, and a minute of arc is 1/60th of a degree.)

How visible would the DS2 (at 160 km diameter) be from Earth at that distance?

Meaning the Moon's distance? The Moon varies between 359,400 km and 402,000 km from the surface of the Earth, making a 160 km object at that distance take a viewing angle of .00040 to .00045 radians. This is 1.37' to 1.53' of arc. Or 4.6% of the Moon's size (29' to 33'), since 160 km is 4.6% of the Moon's diameter).

I think that's large enough to see it as disk with the naked eye. For comparison, Jupiter is around .3 minutes of arc, or 20". (It varies a lot, of course, depending on distance.)

In summary: Divide diameter by distance to get the angle in radians, then multiply by 180/π to get degrees, then multiply by 60 to get minutes.

BTW, if anyone still has Hyperspace access to sw.com, give me a PM. I need a picture of Ralph McQuarrie's early 150km Death Star 1 sketch:

+http://www.starwars.com/hyperspace/memb ... 50613.html