10 Seconds Of Arc = Distance

644.57ft...

That's what I said .

_{The measurement of the tick-mark spacing was only to 3 sig. figs.}

Awesome Everyone!

Thanks for doing this for me.

Joe

Slide rule V calculator.....

I reckon its about 4 olympic size pool lengths.

I know you've already been given the answer, but the workings out are: (assuming the arc distance ~ straight line distance)

10 second of arc = 10sec / (60sec/min * 60min/deg) converting the angle into a fraction of degrees. =~2.78millidegree

Circumference = 2 * Pi * radius

Arc length = angle/360 * Circumference, using the angle is given above.

Flipping it around

Radius = (arc length * 360)/(2 * pi * angle) = 0.375 * 360/3.14 * 0/00278) = 7735 inches =~644.6foots

This link may help.

I get 64.57 feet

i used trig, thus;

0.1875/Tan 5 sec Halved the angle and halved the scale to give a right angled triangle.

You missed out a 4 to the left of the d.p.

(You can also use the small-angle approximation tan(θ) = θ (with θ in radians). It gives the same answer).

Thanks for the reply. I was actually wrong by a factor of 10. I divided 5 seconds by 360 instead of by 3600.

Too late at night or maybe just getting old.

Should've had some ales to blame it on, that would make me feel better in more ways than one.

OK, further math work.

I also make my own telescopes. And the ultimate method to testing them is called the Star Test.

http://starizona.com/acb/basics/using_startesting.aspx

Of course this is not always convient. a real star needs night time, and cloudless nights. Plus if you use any star other than polaris, it keeps moving on you.

So many people like me use an artificial star.

A star for a star test is more or less a infinately small pinpoint bundle of light coming from the sdustant star a m more or less perfectly parallel bundle of light.

Many people like me make a artificil pinpoint bundle of light by using our star "The SUN" But here on earth the sun is NOT a pinpoing of light. so a star test will not work with it. BUT. we make it "Seem" to be a pinpoint of light.

In my case I use a small silver plated glass Christmas ornament placed on a stick up on a nearby hill about 1000 feet away.

The small radius of this ornament makes the sun appear as a pinpoint. But I wonder what actual size it really is,,, So here re the parameters.

The Sun itself in the sky has an angle of arc that averages 30 minutes of arc.

The christmas ornament has a radius of 0.25 inches

And it is placed 1000 feet away.

with this set up, The image of the sun reflected from the ornament, gives a diameter of how many arc seconds?

Joe

Well, here's an attempt ...

AB = distance from observer to reflector = L (1000ft)

BC = radius of reflector = R (0.25in)

HJ is the tangent at point C

D represents one edge of the disk of the sun.

1. Angle CAB is found from tan(CAB) = CF/AF, where CF=√(R²/2)

2. Angle ACB is found using the sine rule: sin(ACB) = sin(CAB) * L/R

3. Angle ECD is found by elementary geometry.

Now add 0.5' of arc to angle ECD, and work backwards to find angle C'AB, where C' is the point on the radius hit by a ray from the other edge of the disk of the sun and reflected back to the observer. The required angle is (C'AB - CAB).

Sorry for the messy description, but I think the method will work (at least as an approximation). I put the figures into an Excel sheet (may have made a mistake or two converting degrees to rads etc.), and I come up with the angle of the image of the sun seen by the observer = 0.027 arc seconds.

I'll run the sums with another drawing, so I can use AutoCAD to find the actual angles, and see whether the method works.

Haven't spotted where I've gone wrong - yet - but my result from the spreadsheet seems to be exactly twice the result from the AutoCAD drawing. That makes your figure 0.013 arc seconds.

Have checked and corrected the spreadsheet and also worked it out by another method. Now I consistently get 0.013 arc seconds (assuming that the sun is directly overhead).

Which is still darn near a pinpoint source of light!