+Graham Posted February 23, 2014 Share Posted February 23, 2014 I have come across an old scope mirror and I need to check its F length. The mirror is 150.00mm in diameter and has a Sagitta of 3mm. Working through using the chord length to get the the radius of the Sagitta curve I make it to have a focal length of 469.5mm or F3.13. Can someone please run through it for me to make sure I have this correct. Thanks. Link to comment Share on other sites More sharing options...
Brantuk Posted February 23, 2014 Share Posted February 23, 2014 F3.13 is correct when you divide aperture into focal length. But the chord maths with the "sagitta" isn't something I'm familiar with. I'll leave that bit to someone more knowledegable. Link to comment Share on other sites More sharing options...
+Graham Posted February 23, 2014 Author Share Posted February 23, 2014 F3.13 is correct when you divide aperture into focal length. But the chord maths with the "sagitta" isn't something I'm familiar with. I'll leave that bit to someone more knowledegable. Thanks Kim Using the chord I calculate the radius of the Sagitta to be 939mm. Focal length I believe to be half of this so it will be 469.5mm Or I could be completely wrong Link to comment Share on other sites More sharing options...
Perkil8r Posted February 23, 2014 Share Posted February 23, 2014 focal length will be 939mm. Useful linkage: http://www.mathopenref.com/sagitta.html Radius will be = to focal length or near as damn it. The focal point is the point at which any tangents (wrong word, can't think what it is though, it may be cord?) to the mirrors surface will converge. This will therefore naturally be the centre of the circle, giving the radius as the distance from mirror face to the centre. Link to comment Share on other sites More sharing options...
+Graham Posted February 23, 2014 Author Share Posted February 23, 2014 focal length will be 939mm. Useful linkage: http://www.mathopenref.com/sagitta.html Radius will be = to focal length or near as damn it. The focal point is the point at which any tangents to the mirrors surface will converge. This will therefore naturally be the centre of the circle, giving the radius as the distance from mirror face to the centre. Cheers Mike For some reason I thought the focal length was half the radius. So that would make it F6.26 Link to comment Share on other sites More sharing options...
Perkil8r Posted February 23, 2014 Share Posted February 23, 2014 No worries Graham. that does also answer your previous questions regarding focal ratios. Remember that light from distant objects can be considered as being in parallel lines, so by reducing the column of light hitting the mirror (by cropping the open end of the tube) it has the same effect as reducing the dia of the mirror to the same. Therefore the figure of the mirror remains the same, as such the focal length doesn't alter, but fl/dia does, hence focal ratio changes. Link to comment Share on other sites More sharing options...
+Graham Posted February 23, 2014 Author Share Posted February 23, 2014 I have just been given a link to a site which works it all out for you. So much easier that churning over the maths. https://stellafane.org/tm/atm/mirror-refs/sag2fl-calc.html Might be a handy link to keep hold of. Link to comment Share on other sites More sharing options...
Perkil8r Posted February 23, 2014 Share Posted February 23, 2014 That seems counter intuitive but there we go then, seems you were right first time. Link to comment Share on other sites More sharing options...
Johnnyaardvark Posted February 23, 2014 Share Posted February 23, 2014 (edited) This looks as good as any explanation.http://www.physicsclassroom.com/class/refln/u13l3a.cfmA ray of light produced at the centre (radius from the surface of the mirror) of a complete circle/sphere would hit the surface at 90° to the surface (i.e. and angle if incidence of 0°) so would be reflected directly back along that line to the centre. Distant stars producing parallel rays have their light reflected to the focus at 1/2 the radius. No worries Graham. that does also answer your previous questions regarding focal ratios. Remember that light from distant objects can be considered as being in parallel lines, so by reducing the column of light hitting the mirror (by cropping the open end of the tube) it has the same effect as reducing the dia of the mirror to the same. Therefore the figure of the mirror remains the same, as such the focal length doesn't alter, but fl/dia does, hence focal ratio changes. ?? Even through a "perfect" pinhole (effectively zero diameter - assuming zero thickness to the mask) rays could be incident on the full diameter of the mirror. Have to say I am not fully sure about all this with respect to f-ratio. The Sagitta is a new quantity to me! I useful link... may be I should just build my own telescope Only of I had the time! Steve Edited February 23, 2014 by Johnnyaardvark Link to comment Share on other sites More sharing options...
Perkil8r Posted February 23, 2014 Share Posted February 23, 2014 This looks as good as any explanation. http://www.physicsclassroom.com/class/refln/u13l3a.cfm A ray of light produced at the centre (radius from the surface of the mirror) of a complete circle/sphere would hit the surface at 90° to the surface (i.e. and angle if incidence of 0°) so would be reflected directly back along that line to the centre. Distant stars producing parallel rays have their light reflected to the focus at 1/2 the radius. Steve Of course I was forgetting that the light is reflected off t an angle that explains quite concisely why I was wrong in saying it would be at the radius and not half of the radius. Thanks Steve I blame paint fumes Link to comment Share on other sites More sharing options...
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