## Article (Image Capture):

## How to Very Accurately Calculate your Telescope's Effective Focal Length

You might be reading this thinking there is absolutely no point to it since if your manufacturer tells you your telescope's focal length and you use a known focal extender/reducer, you can easily work out the effective focal length. However, sometimes it is not so easy. For example,

Why is this even important? Well, some of us like to image mosaics. When it comes to imaging mosaics, one needs to know the effective focal length of the telescope-camera setup pretty well in order for the field of view to be calculated correctly, which in turn leads to nicely overlapping mosaic image segments.

I would like to start by apologising because I cannot just provide a mathematical equation without its derivation. I hate being given equations and not how they arrive at them. So here it goes, a short derivation of an equation that will give you your telescope's effective focal length with attempted

**Skywatcher**produce their**2" Coma Corrector**for*f/4-f/6*telescopes (optimised for their*f/5*range) and do not state anything about it changing your effective focal length. A quick test and calculation however reveals it is actually effectively a*0.9x*focal reducer as well. Proof of this later. Also, my**Altair Astro 8" RC**telescope is coupled with my**Astro-Physics CCDT67 Telecompressor**and this produces a focal reduction effect depending on your CCD sensor's distance from the telecompressor lens. You can predict it if you know the distance very accurately but I feel it is better to know the effective focal length straight from physical evidence.Why is this even important? Well, some of us like to image mosaics. When it comes to imaging mosaics, one needs to know the effective focal length of the telescope-camera setup pretty well in order for the field of view to be calculated correctly, which in turn leads to nicely overlapping mosaic image segments.

I would like to start by apologising because I cannot just provide a mathematical equation without its derivation. I hate being given equations and not how they arrive at them. So here it goes, a short derivation of an equation that will give you your telescope's effective focal length with attempted

**Microsoft Paint**diagrams to accompany the mathematics.Above is a very, very simplistic diagram of a telescope-camera setup. A light ray enters the OTA and is focused on to the CCD sensor in the CCD camera. The CCD sensor is comprised of bins - the physical

*boxes*where photons of light fall in. As is natural, these bins have a finite size. The size of a pixel in your images is therefore partly assigned by this physical bin size of your CCD sensor and makes the following triangle with the focal length.Above, I call

**FL**the focal length of the telescope,**PS**the pixel size and**θ**is therefore the angle. Simple trigonometry yields the following relationship:tan θ=PS/FL

As you can imagine, my triangle diagram is greatly, greatly exaggerated in scale. A pixel is much, much smaller than what is shown in my diagram. Therefore, the angle

**θ**is tiny in reality. As a trigonometric approximation,**tan θ ≈ θ**so we can actually write the above as:θ=PS/FL

The unit of angle here is the natural one,

*radians*so**θ**is in*radians*. However,**PS**is in distance per pixel (e.g.*millimetres*per pixel) and**FL**is in distance (e.g.*millimetres*). As we are dividing**PS**by**FL**, the distance units cancel each other out and simply leave behind units of*"per pixel"*. As a result:θ=PS/FL[radians per pixel]

i.e. the unit of

**θ**is actually*radians per pixel*. In astrophotography however, we are much more familiar with the unit*arcseconds per pixel*, which is a measure of resolution achieved or achievable. In a circle there are*2π radians*, which is the same as*360°*. This means that*360 / 2π = 57.29578°*per*radian*. In each degree there are*60 arcminutes*and in each*arcminute*there are*60 arcseconds*. That means that in each*degree*there are effectively*60 x 60 = 3600 arcseconds*. So, how many*arcseconds*are there in a*radian*? Simple, that is*57.29578 x 3600 = 206264.8 arcseconds per radian.*This is an important result because we can now say that:θ= (206264.8 xPS) /FL[arcseconds per pixel]

We have multiplied the entire equation (effectively the numerator of this fraction) by the factor of

*206264.8*in order to convert from*radians*to*arcseconds*. We also note that pixel size is usually given in*micrometres per pixel*and focal length is usually given in*millimetres*. There is a*1000*difference between the magnitudes of distance here. We can absorb this*1000*difference into the equation above by dividing our factor of*206264.8*by*1000*:θ= (206.2648 x PS) /FL[arcseconds per pixel]

The quantities

**PS**and**FL**are now in familiar units -**PS**is in*micrometres per pixel*and**FL**is in*millimetres*. In more familiar terms, we can conclude this derivation by this equation:[Arcseconds Per Pixel] = (206.2648 x [Pixel Size in μm]) / [Focal Length in mm]

So where does this equation come in to play? Well, we can use evidence to calculate the effective focal length of your telescope. Set up your telescope and camera how you would use it for imaging (with any focal reducers/telecompressors, spacers, etc) and capture a single exposure of whichever target you like in the night sky. Ensure the focus is good and that you capture a decent exposure with numerous stars across the field of view. For now, do not use binning (leave at

Armed with your image's resulting

*1x1*). Once you have the image, upload it to**Astrometry.net**. This will plate solve the image, i.e. it will figure out what target it is you were looking at as well as label all the recognisable objects across your entire image. The important thing here is that under the*Calibration*information shown, it will give you a very accurate value of*pixel scale*in*arcseconds per pixel*. Please note that if you upload an image that was binned at say*2x2*, the*arcseconds per pixel*will be*twice*as large.Armed with your image's resulting

*arcseconds per pixel*, we can re-arrange the above final equation to:[Focal Length in mm] = (206.2648 x [Pixel Size in μm]) / [Arcseconds Per Pixel]

With this, you will get a very accurate value of your telescope's effective focal length. Again note that if you upload a binned image, you will need to divide your

Here is an example. I used to use a

*arcseconds per pixel*value by your binning amount first (e.g. if you binned at*2x2*, divide your value of*arcseconds per pixel*by*2*before proceeding). From that point on, armed with this information plus your CCD sensor's details (from the CCD camera's manufacturer), you can easily calculate your field of view in sources such as the**Sky at Night Field of View Calculator**or programs like**CCDCalc**. Quite importantly, this will provide you with the ability to accurately determine your field of view for mosaic imaging. Also, since your telescope aperture is obviously not being changed, by dividing your effective focal length by your aperture, you can calculate your effective focal ratio, also very accurately.Here is an example. I used to use a

**Skywatcher Explorer 150PDS**telescope with an**ATIK 383L+**CCD camera. I uploaded a*7 minute*exposure in*Luminance*(along with a**Skywatcher 2" Coma Corrector**and**Hutech IDAS LPS 2"**filter) of the*Elephant Trunk Nebula*region to**Astrometry.net**. I was given a resolution of*1.63 arcseconds per pixel*. According to**ATIK**, my CCD camera has a pixel size of*5.4μm per pixel*, so:[Focal Length in mm] = (206.2648 x 5.4) / 1.63 = 683.331

With my imaging optical train, the effective focal length of the telescope was

*683.331mm*. If you look at the manufacturer specifications of the telescope, it is supposed to have a focal length of*750mm*. A*750mm*focal length along with the*150mm*aperture yields an*f/5*focal ratio but with a focal length of*683.331mm*, the focal ratio is actually*f/4.556*and there is a focal reduction factor of*0.911x*. In essence, despite**Skywatcher**not advertising their*2"**Coma Corrector*as changing anything, it does provide some focal reduction, as evidenced. This is not a criticism of**Skywatcher**or their*2"**Coma Corrector*- personally I thought it was excellent that I was getting a faster optical system as a result of using it and the flatness of the images was fantastic - but it should give you an idea of the importance of calculating your effective focal length with your imaging optical train. Estimating it may be sufficient for single images but for mosaics, it is a good idea to know precisely what you are working with beforehand. Comment Box is loading comments...