DIMENSIONS OF A NEWTON TELESCOPE
We suppose that: D (diameter of the main mirror), f (distance of focus to optical axis),
F (focus of mirror) en d (diameter of desired illuminated field) are known. Then:
g/d=(g+F)/D --> g=Fd/(D-d) and (g+f)/b=(g+F)/D --> b=D(g+f)/(g+F)
(b is a fair approximation of the size of the secondary mirror; error is h1+h2-b) and:
y=x en y=-dx/2g+1/2b --> h1=gb/(d+2g) and L1=h1*SQRT(2), likewise:
y=x en y=dx/2g-1/2b --> h2=gb/(2g-d) and L2=h2*SQRT(2)
L1+L2 being the total length (long axis) of the secondary mirror.
The short axis is (L1+L2)/SQRT(2)
The center of the sec. mirror should be (h2-h1)/2 lower than optical axis.
more direct formulae:
h1=d/2 + (D-d)(f-h1)/2F --> h1= (dF+(D-d)f)/(2F+(D-d))
h2= d/2 + (D-d)(f+h2)/2F --> h2=(dF+(D-d)f)/(2F-(D-d))
h=h1+h2, i.e. the short axis of the secondary mirror
L=L1+L2 is the long axis, the length of the secondary mirror
field of view (FOV) is in arcminutes: 60*s/(F*tan(1)),
where s is the diameter of the field and tan(1) is in degrees, so 0.017455065 (about 1/57.3).
The s for ccd-cameras is the SQRT(length of chip^2+width of the chip^2), e.g for the ST-7: 8.29 mm, i.e. the length of the diagonal.
However this can be limited by the inner width of the tube (I).
Supposing the main mirror is centered in the tube than the maximum FOV is: 2arctan((I-D)/2M)
An example: D=100, I=150 and M=500, then the FOV is 2arctan(0.05), meaning "find an angle in case where the tangent is 0.05 and multiple this by two", so 5.7 degrees.
We also can derive d from D, f, F and h directly.
h=h1+h2=gb/(2g+d)+gb/(2g-d)=Fdb/(2Fd+dD-d)+Fdb/(2Fd-dD+d)
as b=D(g+f)/(g+F)=(Fd+fD-fd)/F:
h=(4F^2d+4fFD-4fFd)/(4F^2-D^2+2dD-d^2) -->hd^2-2hdD+4dF^2-4fFd-4hF^2+hD^2+4fFD=0
hd^2+(-2hD+4F^2-4fF)d-4hF^2+hD^2+4fFD=0
from this: according to the generic formula x=(-bħSQRT(b^2-4ac))/2a
d=(2hD-4F^2+4fF+SQRT((2hD-4F^2+4fF)^2-4h(-4hF^2+hD^2+4*fFD)))/2h
If you need a small, simple spreadsheet (Excel 4.0) with the above calculations, just send me an e-mail.
(deepsky [at] hetnet.nl)