Telescope alteration (Russ or turbo-1)

[Saladsamurai]

I'll take any information, I just know that Russ and Turbo have helped me before.

Okay, here is the rundown: I have a really inexpensive ($150) Meade reflector telescope.

Here are the specs: 114 mm primary, F=1000mm, f/8.8

I am not quite sure what the F=1000mm, f/8.8 means.

Now, the telescope is a little strange in that it has some sort of 'lens' located in the focuser tube. Yes a lens. It is midway between where the eyepiece is inserted and where the focuser tube ends in the optical tube.

It serves as some sort of relay...or a way of allowing them to make the telescope shorter and more compact. (Russ, you may recall this)

I really feel like this lens is taking away from the viewing experience. I feel it worsens the telescopes optics.

I wouldn't mind experimenting with the scope. I have an extra Crayford style focuser lying around. I was thinking of removing the old 'relay' focuser', cutting a larger hole in the optical tube, and putting in the Crayford.

I am just not sure how this will change things. I am sure there will be some math involved, but I don't mind.

Will this change the focal length? What does that mean anyway? Will I need to Adjust how high the focuser assembly is from the optical tube? (I assume I will, though I can't exactly say why!) If so, how can I figure out how high?

I know this is a lengthy question and there are other astronomy sites where I could ask. But, I am convinced that I receive the best information here. And I am willing to expend the energy to learn how to figure these things out. I think this experiment will help me get familiar with the parts and workings of the telescope so that when I attempt to build one this summer I will at least know some basics.

Thanks

 

[mgb_phys]

The f=1000mm is the focal length of the telescope, it's simply the distance from the mirror where parallel light would come to a focus.

The f8.8 is the focal ratio = focal length/aperture = 1000/146
This is a measure of the brightness of the image and the field of view,
A larger aperture collects more light and a faster (smaller f number) telescope gives a larger field of view. But a faster mirror is more steeply curved and so more difficult and expensive to figure.

I don't know this telescope but I'm guessing that if it is a short tube then the position of the focal plane 1000mm from the mirror would be too far outside the tube to mount an eyepiece and so the relay lens relocates it at a more convenient point - but possibly at the cost of some field of view.

 

[Saladsamurai]

The f=1000mm is the focal length of the telescope, it's simply the distance from the mirror where parallel light would come to a focus.

The f8.8 is the focal ratio = focal length/aperture = 1000/146
This is a measure of the brightness of the image and the field of view,
A larger aperture collects more light and a faster (smaller f number) telescope gives a larger field of view. But a faster mirror is more steeply curved and so more difficult and expensive to figure.

I don't know this telescope but I'm guessing that if it is a short tube then the position of the focal plane 1000mm from the mirror would be too far outside the tube to mount an eyepiece and so the relay lens relocates it at a more convenient point - but possibly at the cost of some field of view.

This makes sense. The tube is short...let me see if I can find a pic. Here it is...it's the shorter one

So since F=1000mm=100cm is most likely outside of the telescope the lens must do something to correct that.

Now it seems to me that the focal length must be a function of the degree of curvature of the primary mirror. So, is there anyway to get around using the lens? Or am I better off just making a new optical tube?

 

[russ_watters]

I'm thinking it serves as a barlow and that if you remove it, you may just be left with a shorter focal length scope, which can still focus at a reasonable distance from the tube.

 

[turbo]

If you can remove the intermediate lens without damage, and retain the ability to re-install it, try that. Remember that the lens might not be symmetrical, so mark it so you'll know which surface of the lens goes toward the eyepiece. Another thought - that intermediate lens could serve as a field flattener instead of a barlow, so that the whole field of view comes to focus at about the same EP distance. I'm not familiar with this scope, but I'll dig around a bit.

 

[mgb_phys]

I was thinking that - but why would you make a fast reflector and then slow it down to f8.8 ?
Wouldn't the only advantage of a slower telescope be a cheaper primary and a smaller eyepiece?

 

[turbo]

After a little research, I think Russ is right, and that intermediate lens is a Barlow. The configuration of the Meade 114EQ-ASTR is similar to that of the Orion Star-Blast 4.5" and it is marketed as a 450mm f/4. If you remove the Barlow, you will only lose magnification. The trade-off is that you will have a much larger field of view for observing extended objects like nebulae and the galaxies in our Local Group, AND everything will be brighter. At the $150 price-point any built-in Barlow is not likely to be optically great, nor is it likely to be fully coated to suppress reflection, so you may indeed improve the optical performance of the scope by removing it.

 

[Chronos]

That extra optical component is called a relay lens. It significantly shortens the focal distance allowing for a shorter tube and less bulky mount.

 

[Saladsamurai]

Okay, so here is what I am thinking. I already bought another much nicer telescope, so I would really like to mess with the components of this one.

I could just remove the relay lens and end up with a shorter focal length.

Or (and this is what I would like to do and probably will)

Retain the original long focal length by removing the primary and secondary and use them with my spare Crayford to make a Truss-tube style scope.

I think this would be a great preliminary to building my own scope from scratch. I could least get a sense of how to design the housing and its components.So since the focal length is 1000mm, that means that when I design the housing, I should set it up so that the secondary will lie approximately 1000mm from the primary right?

Is that how it works?

Thanks for all of the responses too guys!

 

[turbo]

Making your own truss tube and transferring components is a great way to learn. Remember that the 1000mm is the length of the light-cone from the primary to the focal point. Your secondary needs to be closer than 1000mm to the primary to allow for the length of the focuser - the EP needs to be 1000mm from the primary. As you build your truss-tube you may want to consider making a design in which the secondary and the focuser can be adjusted to be closer to or farther from the primary, as needed. With some EPs, accessories, cameras, etc, you may need more in-focus or back-focus ability that you originally anticipated.

 

[Saladsamurai]

so the 'eyepeice' has to be 1000mm from the primary mirror? I don't know why that confuses me.

If the EP is perpendicular to the optical tube and set back from it, how do you take that measurement. I will try to find a diagram. That would make it a lot easier!

 

[turbo]

The 1000mm is the length of the light-path, including the path from the primary to the secondary, and the path from the secondary up the focuser tube to the EP. Different EPs will come to focus at different focuser extensions, so you should expect to plan for that.

 

[Saladsamurai]

Ohhhh. I get it. 1000mm is the TOTAL length from the primary to the EP. Since it is a reflector, that length is in a sense being 'bent' around a corner since you cannot put the EP where it would block the incoming light to the primary.

It all makes sense now. Since the secondary is flat, it does the hinder the light as it comes to a focus... it only acts to redirect said light.

 

[Saladsamurai]

I have two more quick questions. First,I am still not 100% on this.

The f8.8 is the focal ratio = focal length/aperture = 1000/146
This is a measure of the brightness of the image and the field of view,

How do I use this "f" number to describe something, that is what does it describe and how? mgb said "This is a measure of the brightness of the image and the field of view"

...but I am not sure how it does that?

And second:

I have mad a collimation mark on my primary mirror.
Why does the light that hits the center of the primary not get reflected to the secondary?

And is there a way to calculate the radius of the surface that will not get reflected?

 

[mgb_phys]

Sorry I didn't phrase that very well -
The aperature is what matters for light gathering power.
The focal length determines the magnification (along with the eyepiece) and the field of view.
The focal length isn't usually quoted, what is given is the aperture and f number - so you can calculate the focal length. So in those terms for a given aperture the f number determines the focal length, hence the magnification, hence the brightness and fov.

This is a good description of some of the optics http://www.televue.com/engine/page.asp?ID=140


Collimation - this page seems to make sense http://web.telia.com/~u41105032/kolli/kolli.html

 

[Chronos]

The explanation by mgb is correct, but perhaps a bit technical. A short focal length [small f number] objective has less intrinsic magnification than a longer focus objective. Lower magnification means images are smaller, but brighter, and you can see a larger chunk of the sky. BTW, removing the relay lens from your scope will significantly increase the focal distance. The whole point of the relay lens is to reduce manufacturing costs [shorter tube, smaller mount].

 

[RonL]

http://www.bbastrodesigns.com/
This link leads to Mel's homepage, go to telescope making, and almost everything is covered.

I was glad to see Mel Bartel's page mentioned at the bottom of the page you linked to, about Collimation.

 

[Saladsamurai]

The explanation by mgb is correct, but perhaps a bit technical. A short focal length [small f number] objective has less intrinsic magnification than a longer focus objective. Lower magnification means images are smaller, but brighter, and you can see a larger chunk of the sky. BTW, removing the relay lens from your scope will significantly increase the focal distance. The whole point of the relay lens is to reduce manufacturing costs [shorter tube, smaller mount].

Makes sense. Now as for the lens. I plan on removing that. I actually plan just using the mirrors and EP's from that scope along with a real focuser (no relay) and a truss assembly.

Now the focal length has nothing to do with the lens I thought. I is an inherent property of the curvature of the mirror. So why would the focal length change. It will still be 1000mm won't it? That is, I thought the relay was just a way of handling such a long focal length in such a short tube. 1000mm?Maybe it would be easier for me to just say: Forget about the relay. If I take out the mirrors altogether and that is all I salvage from the scope, what kind of focal length will I be looking at in order to design a new tube assembly?

 

[turbo]

Now the focal length has nothing to do with the lens I thought. I is an inherent property of the curvature of the mirror. So why would the focal length change. It will still be 1000mm won't it? That is, I thought the relay was just a way of handling such a long focal length in such a short tube. 1000mm?

It would be more accurate to say that with the lens, the effective focal length is 1000mm. When you remove that lens the actual focal length (which IS a property of the curvature of the mirror) should be shorter - perhaps in the 400-500mm range. You'll have to measure that focal length and subtract the average focus tube extension to determine the how far the secondary mirror needs to be from the primary.

 

[Saladsamurai]

Ahh. How would I 'measure' the acual focal length? Is there a simple way to do it?

 

[turbo]

Here is a method, and it goes on to explain how to set up the mirror/focuser in a Newtonian design. Google will get you to all kinds of ATM sites with good info.

http://www.astro.lsa.umich.edu/undergrad/Labs/optics/Reflectors.html

 

[Saladsamurai]

Here is a method, and it goes on to explain how to set up the mirror/focuser in a Newtonian design. Google will get you to all kinds of ATM sites with good info.

http://www.astro.lsa.umich.edu/undergrad/Labs/optics/Reflectors.html

Seems like you need a bit of equipment for this. There must be an easier way to do it with household objects.

thanks turbo, I will google around some more.

 

[turbo]

Wax paper is translucent and you could tape a sheet of that to a frame of some sort for you projection screen. If you mount the mirror vertically on a table top with the screen parallel to the mirror, but offset so that some light from a distant object will bypass the screen, you should be able to slide the screen closer or farther until you get a focused image on the screen. The distance between the mirror and the screen is the focal length of the mirror.

 

[tony873004]

You may notice that the 1000mm is labeled as "effective focal length", not "focal length". Get a ruler and measure the distance from the primary to the secondary, then add roughly half the tube diameter. You'll probably get around 500 mm. The built-in Barlow increases this by about 2x, hence an "effective focal length" of 1000mm.

I think I have the same telescope as you, and I was confused as to why the focal length was 1000 mm, but the telescope was only about half this length. So I called Meade Technical support, and that's what they told me. Prior to that, I didn't even realize that there was a piece of glass in the focuser.

 

[russ_watters]

The short answer to the focal ratio question is that since area is a square function of aperature, cutting the focal ratio in half by cutting the focal length in half (from, say, 1000mm to 500mm: f/8 to f/4), gives you an image 4x as bright, and with half the magnification (or twice the field of view).

 

[Saladsamurai]

Wax paper is translucent and you could tape a sheet of that to a frame of some sort for you projection screen. If you mount the mirror vertically on a table top with the screen parallel to the mirror, but offset so that some light from a distant object will bypass the screen, you should be able to slide the screen closer or farther until you get a focused image on the screen. The distance between the mirror and the screen is the focal length of the mirror.

This is similar to something that a friend and I were discussing. I like it. Now by offset, do you mean something like what I have drawn below? this is a view from above. /---------\ <---mirror
|
|
|
--------- <----screen


X

"X" marks the placement of an object, maybe a few feet away from the screen. like a lamp or something.

Edit: Goddamnit! when I submit post, the picture changes!

Hit QUOTE to see what it actually looks like

 

[mgb_phys]

cutting the focal ratio in half by cutting the focal length in half (from, say, 1000mm to 500mm: f/8 to f/4), gives you an image 4x as bright, and with half the magnification (or twice the field of view).

That's what I didn't understand - you make an 146mm f/4 and then slow it to f9 and all you gain is extra useless magnification. Especially since it must be much harder to figure an f/4 mirror than a f/9

 

[russ_watters]

That's what I didn't understand - you make an 146mm f/4 and then slow it to f9 and all you gain is extra useless magnification. Especially since it must be much harder to figure an f/4 mirror than a f/9

Two reasons:

1. Advertising.
2. Newbies mostly want to view planets.

 

[turbo]

Casey, your drawing is not far off the mark. You can allow the screen to obscure half of the mirror and still get a nice focused image on the screen, just try to use a distant object. If you can get a decent image of a distant building or tree-line, that should be close enough.

 

[Saladsamurai]

Righto! I will try to try it tomorrow weather permitting (stupid New England!).

Thanks y'all. I'll let you know how it turns out.

 

[Chronos]

It would be more accurate to say that with the lens, the effective focal length is 1000mm. When you remove that lens the actual focal length (which IS a property of the curvature of the mirror) should be shorter - perhaps in the 400-500mm range. You'll have to measure that focal length and subtract the average focus tube extension to determine the how far the secondary mirror needs to be from the primary.

You have it backwards, turbo-1. The primary mirror is figured at f8.8 [which is much cheaper than say f4.4]. Removing the relay lens will significantly increase the focal distance of the telescope.

 

[Saladsamurai]

So, it will be longer. Well that's cool. I already have a larger scope with an f4.9.
So the longer focal lengths are usually better for planetary viewing, right?

 

[Chronos]

The intrinsic magnification is greater, hence you will get more magnification from a less powerful eyepiece. For planetary viewing, this is an advantage. Short focal length eyepieces have a narrow exit pupil [more difficult to use].

 

[turbo]

Righto! I will try to try it tomorrow weather permitting (stupid New England!).

Thanks y'all. I'll let you know how it turns out.

I'm in stupid New England, too (central Maine) and we had 40-50 mph winds almost all day. Luckily, they were fairly consistent and not gusty, or the resonances in the gusts would have weakened/toppled trees. Amazingly, our power stayed on all day. Yay!

 

[chemisttree]

So, it will be longer. Well that's cool. I already have a larger scope with an f4.9.
So the longer focal lengths are usually better for planetary viewing, right?

No, Turbo had it right. It is a short focal length telescope with an integral barlow yielding an effective focal length of 1000 mm. Remove the barlow and the effective focal length will be lower.