The speed objects move accross your eyepiece

[jabeoo1]

I have a question about how long it takes an object to move from one side to the other side of your field of view on an unmounted telescope at a given magnifications, I do not own a scope so cannot test my maths against the reality.  I appreciate anyone who can let me know of errors in my method:

The following example:

If a 40mm eyepiece with apparent field of 43 degrees in a scope gives x 40 magnification and a true field of 1 degree. 

The movement of the sky to us appears to drift 15 degrees / hour.  That makes 0.25 degrees / minute.  So does that give you 4 minutes of observing time across the 1 degree of true field ?

Or not?

Thanks in advance.

James

[ScottS]

Surely it depends where in the sky you are looking. The way I understand it, you could point your scope at Polaris and look at it all night without moving the scope.

[Moonshane]

yes - spot on. that's a really easy way to think of it actually. 0.25 degrees per minute.

[Moonshane]

Polaris is a special case as are things close to it that obviously revolve around it. they will still move across 15 degrees per hour but at the pole they are 'bunched closer'. it's a good point though and I suppose near Polaris the eyepiece also covers more 'sky field' for the same reason. never thought about that.

[arrayschism]

If you are interested, Ryukyu Astro Club's home page has a neat (declination dependent) formula to calculate this, as long as you know your TFoV:

http://www.nexstarsite.com/_RAC/form.html#22

[jabeoo1]

To extend the concept :  If you were to x 2 barlow the same example to x80 magnification will that divide the true field of view so it becomes 0.5 degrees ?  Or not. 

Thanks in advance again. 

[Moonshane]

yes, true field is roughly apparent field of the eyepiece / magnification.

[Scooot]

If you are interested, Ryukyu Astro Club's home page has a neat (declination dependent) formula to calculate this, as long as you know your TFoV:

http://www.nexstarsite.com/_RAC/form.html#22

This is interesting. I've been trying to work out the transit time of Betelgeuse over 1 degree. Can anyone tell me what Abs is in the transit time formula.

[ScottS]

I think Abs is absolute value as its the same for either positive or negative declination.

[Scooot]

I think Abs is absolute value as its the same for either positive or negative declination.

Thanks. I'm getting roughly 1.75 mins but I think its wrong. Would have expected nearly 4 mins as it's near the horizon. I've taken on more than I can chew with this formula. 1*240*(cos 7.4) /60 ?

[Scooot]

Thanks. I'm getting roughly 1.75 mins but I think its wrong. Would have expected nearly 4 mins as it's near the horizon. I've taken on more than I can chew with this formula. 1*240*(cos 7.4) /60 ?

Ah solved it! My cosine calculator was on RAD. It is 3.96 mins. You can tell its cloudy.

[acey]

If a 40mm eyepiece with apparent field of 43 degrees in a scope gives x 40 magnification and a true field of 1 degree. 

The movement of the sky to us appears to drift 15 degrees / hour.  That makes 0.25 degrees / minute.  So does that give you 4 minutes of observing time across the 1 degree of true field ?

The initial mistake was to confuse degrees of declination with degrees measuring apparent field. Imagine the celestial sphere projected onto a flat surface (a photograph) with Polaris at the centre. Then stars would make tracks, and if you measure the angle each of these tracks makes with the centre in one hour you'd find them all to be about 15 degrees. But stars further from the centre make longer tracks, i.e. they cover more sky and would pass more quickly through the field of view of a telescope. The factor for this is cos(declination). On the celestial equator (declination = 0) the factor is 1, while at the celestial pole (declination 90 degrees) it is zero.

Stars actually appear to make a full revolution of the sky in just under 24 hours (because of Earth's motion around the sun): the sidereal day is 86,164 seconds. Then for an eyepiece with true field of view TFOV you get the formula:

Drift time (in seconds) = TFOV x 86,164 / (360 x cos(declination)).

It doesn't matter if the declination is positive or negative since cos(-x) = cos(x).

Betelgeuse has declination 7.4 degrees. To find how long it takes to cover 1 degree of sky, put FOV = 1 and declination = 7.4 to get drift time = 237 seconds, or just under 4 minutes.

[acey]

The formula at Ryuky Astronomy, linked in an earlier post, is wrong. The formula is stated as:

t = TFOV * 240 * Cos(Abs(δ)) [sec]

The 240 comes from using the approximation 24 hours rather than 23 hours 56 minutes (86,164/360 = 239.34), and Abs is unneccessary. The error is multiplication by the cos term, rather than division. For an object at or near the horizon (e.g. Betelgeuse) the cos term is approximately 1 so the error isn't noticeable, but for an object at the pole (e.g. Polaris) it would imply a drift time of zero.

The formula I gave in my previous post (which I can no longer edit) is the correct one.

[Scooot]

The formula at Ryuky Astronomy, linked in an earlier post, is wrong. The formula is stated as:

t = TFOV * 240 * Cos(Abs(δ)) [sec]

The 240 comes from using the approximation 24 hours rather than 23 hours 56 minutes (86,164/360 = 239.34), and Abs is unneccessary. The error is multiplication by the cos term, rather than division. For an object at or near the horizon (e.g. Betelgeuse) the cos term is approximately 1 so the error isn't noticeable, but for an object at the pole (e.g. Polaris) it would imply a drift time of zero.

The formula I gave in my previous post (which I can no longer edit) is the correct one.

Yep I've been thinking about this all day and I'm embarrassed to say my second calculation was wrong as well. I found this page as a good way of understanding it.

http://www.rpi.edu/dept/phys/observatory/obsastro1b.pdf

I've understood it as follows. A star on the celestial equator appears to move 1 degree in 4 minutes, ie a quarter of a degree per minute. Any star that's nearer the pole appears to move proportionately slower. So if its half way nearer it will move half the distance in the same time. A star with a declination of 60 deg is positioned roughly half way because the cosine of 60 deg is about 0.5. So all you need to do is look up the cosine of the star's declination and multiply the distance covered on the celestial equator.

So using my example of Betelgeuse, the declination is 7.4, cosine is 0.9917. So it will move 0.9917 of a degree in 4 minutes.

Does this make sense or have I misunderstood again.

[acey]

Yes, though I think the cosine term is itself an approximation. You could imagine a star very close to the celestial pole, i.e. declination not quite 90 degrees, and an eyepiece with sufficient FOV so that the star is seen to go round and round, never leaving the field. The formula with the cosine term would give a finite drift time whereas the actual time would be infinite.

An exact formula would involve spherical trigonometry. I think the formula I've given is based on the assumption that the path of the star across the field is effectively a straight line, which should be a good enough approximation for stars sufficiently far from the pole.

[Scooot]

Yes, though I think the cosine term is itself an approximation. You could imagine a star very close to the celestial pole, i.e. declination not quite 90 degrees, and an eyepiece with sufficient FOV so that the star is seen to go round and round, never leaving the field. The formula with the cosine term would give a finite drift time whereas the actual time would be infinite.

An exact formula would involve spherical trigonometry. I think the formula I've given is based on the assumption that the path of the star across the field is effectively a straight line, which should be a good enough approximation for stars sufficiently far from the pole.

Thanks acey, yes I understand the round and round near the poles now, but as you say from a practical viewpoint it doesn't matter because there's hardly any drift whilst viewing.

Then as a rough rule of thumb it's fairly easy to remember the following:

Star Drift in Degrees over 4 minutes.

Zero declination 1.00°

30 degrees declination 0.87°

60 degrees declination 0.50°

75 degrees declination 0.25°

I suppose it's another reason why viewing high up with a dob is better, as well as better seeing the object takes a lot longer to drift across the field of view.

[JamesF]

I suppose it's another reason why viewing high up with a dob is better, as well as better seeing the object takes a lot longer to drift across the field of view.

Where "high up" means "closer to the celestial poles" rather than "closer to the zenith".  At the equator, objects at the zenith will appear to move fastest of all.

James

[Scooot]

Where "high up" means "closer to the celestial poles" rather than "closer to the zenith".  At the equator, objects at the zenith will appear to move fastest of all.

James

Good point