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StarFiveSky

Is the Focal length (mm) the tubos' length?

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Always wondered what exactly is the focal length?

For example 114/1000mm

114mm   → Aperture

1000mm → Focal length

So is the telescope a meter (in this case) long in terms of length?

How would you even transport that? If that's the case why not go down a little and get a 114/650 / 114/500 which is way more portable and can even be fit into a backpack.

Or is there a misunderstanding?

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The focal length is the total distance from the primary mirror to the point of focus - in your case out through the focuser in the side of the tube!

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You think you've got problems? 🤣🤣🤣

Yes, roughly speaking the focal length is the same as the tube length, minus the distance from the secondary out to the focuser. This distance gets larger with aperture, so the larger scopes are a little bit shorter than their focal lengths.

This one is 1600mm focal length.

20190710_195813.jpg

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Some optical designs fold the light back and forth within the scope tube before directing it through the focuser. These designs have overall tube lengths quite a lot shorter than the focal length of the primary mirror or obective lens.

 

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Reflectors that have a longer quoted focal length (e.g. 1000mm) than the actual length of their tube (e.g. 500mm) use a lens system before the focuser. These are Bird-Jones type reflectors and are not recommended for beginners: I don't have direct experience of one but from what I've read, they are very difficult to collimate. 

I have a 114/900 f7.9 reflector (tube is 800mm long) - because of its length, it needs a stable  Equatorial mount but gives very good (comparatively narrow) lunar and planetary views - the longer focal length allows for higher mags to be achieved. Another advantage is that it holds its collimation well.

Edited by Peter_D
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Another one I don't kinda get.

Imaging time is directionally proportional to FL.  So why, with my Meade 14" and its vast light gathering power, does an exposure at F10 take so much more than my puny 2.7" refractor at F5????  The light gathering power of the 14" mirror is vastly bigger so is capturing hugely more photons.  So why isn't an exposure quicker with that?

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Hi Kirkster501,

Using a camera of say 5µm pixels your Meade at 3550mm focal length has an image scale of 0.29"/ pixel so each pixel sees 0.29^2 = 0.08 square arcseconds of sky.

The same camera on your refractor at 350mm focal length has an image scale of 3.03"/pixel so each pixel sees 3.03^2 = 9.1 square arcseconds of sky.

Each pixel on your refractor sees 9.1/0.08 = 113 times as much sky compared to the Meade so other things being equal would receive 113 times as many photons.

However the Meade at 355 mm diameter compared to the refractor at 68mm diameter receives (355^2) / (68^2) = 27.2 times as many photons.

Each pixel on the refractor therefore receives 113/27.2 = 4.1 times as many photons as the same sensor on the Meade so needs about a quarter of the exposure time of the Meade for the same image brightness on the sensor. The image on the refractor's sensor will of course be far smaller that that on the Meade's sensor so will show far less detail.

I think that's right unless my sums have let me down. :smile:

Alan

Edited by symmetal
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The length of the tube very roughly approximates to the focal length on some designs and not at all on others. Best to stick with Steppenwolf's clear, simple and correct definition in the first reply.

3 hours ago, kirkster501 said:

 

Imaging time is directionally proportional to FL.  So why, with my Meade 14" and its vast light gathering power, does an exposure at F10 take so much more than my puny 2.7" refractor at F5????  The light gathering power of the 14" mirror is vastly bigger so is capturing hugely more photons.  So why isn't an exposure quicker with that?

No it isn't. It isn't proportional in any way whatever to focal length! And it is only proportional to focal ratio in certain circumstances. Exposure time is proportional to focal ratio in camera lenses where the focal length of the lens is fixed and the diaphragm varies the aperture. The area of unobstructed aperture doubles or halves with each change of F stop. That's why exposure is proportional to F stop (because F stop is proportional to aperture in this case.)

However, if you change the focal ratio by shortening the focal length you are not changing the aperture so you get no more or less light. You do change the number of pixels you put it on, though. So let's think about your 14 inch versus your 2.7 inch refractor. You are mistaken in thinking the refractor is faster. The Meade is, if taking the same object, far faster. However, the object is going to have to be quite small to fit on the chip in the Meade, so let's imagine a small, faint planetary which you intend to present at the same screen size as it would appear in the refractor. Firtly you'll probably be able to bin the camera 4X4 in the Meade and still get the same resolution as you get in the refractor. And then you may still be able to software bin the result and end up with an object the same size as it appears in the refractor image.

Take an image at the same exposure in both, but unbinned in the refractor. The Meade has put about 22 times as much light onto the same number of effective pixels. The little planetary will be far, far brighter in the Meade image. What disguises this fact is that, unless you bin/software bin/resample the capture the Meade will be producing a far larger and far fainter image, but try resampling the refrator's image up to the size of the Meade's and see what it looks like! It will look like noise...

Olly

Edited by ollypenrice
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1 hour ago, johninderby said:

And this is a Skywatcher 180 mak with a 2700mm focal length.

8FF0886F-24A6-4F8B-9583-5A865988B73E.jpeg

I'm sorry but now I'm confused 2700mm = 2.7m

Either your couch is gigantic or I completely lost touch with reality and the replies thrown in here😀

Please explain

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It’s to do with the light path being folded and the shape of the mirrors which means the effective focal length is much greater than the physical length of the telescope.

Quote from an article on the subject.

“At first it seems that an Maksutov-Cassegrain should have a focal length twice as long as the length of the tube, since the light travels twice the length of the tube.  However, Mak-Casses actually have focal lengths about 5-7 times longer than the tube length.  This is due to the convex curvature of the secondary mirror.  This effectively magnifies the focal length, making the scope act much longer than it is “

6151CD8D-5B8B-4976-8F43-496665DC0670.jpeg

Edited by johninderby
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1 hour ago, ollypenrice said:

The length of the tube very roughly approximates to the focal length on some designs and not at all on others. Best to stick with Steppenwolf's clear, simple and correct definition in the first reply.

You are totally correct of course Olly. I somewhat assumed we were talking about newtonians in my reply, but of course even in that case the situation is confused by Bird-Jones scopes, let alone Maks and SCTs etc.

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3 hours ago, johninderby said:
3 hours ago, johninderby said:

It’s to do with the light path being folded and the shape of the mirrors which means the effective focal length is much greater than the physical length of the telescope.

6151CD8D-5B8B-4976-8F43-496665DC0670.jpeg

Oh, I think I understand it now.

Thanks everyone for taking the time to reply and comment!

regards

Edited by StarFiveSky
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