Understand the relationship
The reasoning behind the result
Circular collecting area grows with diameter squared
Agross=πD²/4; Aclear=π(D²−d²)/4
The entrance aperture diameter D defines a circular area. A concentric circular obstruction of diameter d removes its own area. The remaining clear area is the difference of these two circles. Doubling both linear dimensions multiplies each area by four.
The calculation assumes a uniform unobstructed annulus outside the entered central obstruction. Spider vanes, baffle losses, off-axis clipping, uneven illumination and additional masks are excluded. A completely blocked aperture has zero clear collecting area, not an undefined geometric result.
Diameter percentage is not area percentage
fblocked=(d/D)²; d=D√fblocked
An obstruction spanning 30% of the aperture diameter blocks 0.30²=0.09, or 9%, of the circular area. Treating 30% diameter obstruction as 30% area loss would overstate its effect on geometric collection.
The three forward input modes accept diameter, diameter percentage or area percentage and reconcile them into the same ledger. These quantities describe collecting area only. Central obstruction also changes diffraction and contrast behavior, which cannot be replaced by this area subtraction or by an equivalent clear diameter.
Transmission weights the light collected by the clear area
Aeffective=T Aclear; R=Aeffective/Aeffective,reference
Entered optical transmission T is a fraction from zero to one. Multiplying clear area by T gives an equivalent collecting area for the specified conditions. Dividing it by the similarly defined reference gives the relative collected flux for the same source and matched spectral conditions.
Transmission does not physically reduce the aperture diameter. The equivalent unobstructed, perfectly transmitting diameter is only a compact representation of the same collected light. It does not reproduce the original instrument's angular resolution, point-spread function, field of view or brightness perceived through an eyepiece.
An inverse ratio fixes a required area under entered losses
Drequired=√[4R Aref/(πT(1−q²))], q=d/D
For a chosen positive collection ratio R, the reference effective area determines the required effective area. The entered obstruction diameter fraction q and transmission then determine the aperture diameter that supplies that area in this model.
Zero transmission or complete obstruction cannot achieve a positive target at any finite aperture and is rejected in the inverse mode. This mathematical requirement does not establish that the aperture, structure or optical design is practical. Actual exposure time and detection performance require a noise and detector model.
Follow the numbers
A 30% obstruction and 80% transmission
- Take a 200 mm aperture with a 60 mm obstruction. The diameter fraction is 60/200=30%, while blocked area is 9% and clear area is 91% of the full aperture.
- At 80% transmission, effective collecting area is 0.91×0.80=0.728 of an unobstructed, perfectly transmitting 200 mm reference.
- The area-equivalent clear diameter is 200√0.728≈170.645832 mm. It represents the same collected light under the entered assumptions, not the same diffraction or contrast.
Obstruction and transmission losses are distinct, and both remain visible in the area ledger.
Quick guide
How to use this calculator
- Define the reference aperture, obstruction and transmission under the same light conditions as the scenarios.
- Select whether obstruction is a diameter, a percentage of aperture diameter or a percentage of aperture area.
- Enter named scenarios or a positive desired collection ratio for the inverse solve.
- Read geometric area and optical throughput separately. Equivalent collecting diameter is not equivalent resolving power or perceived image brightness.
Calculation method
Calculation and interpretation
Keep aperture diameter, blocked area, optical transmission and reference-relative light collection separate.
Aclear=π(D²−d²)/4; Aeffective=T·Aclear; collection ratio=Aeffective/Aref; area obstruction=(d/D)²; Dequivalent=D√[(1−(d/D)²)T].
Worked example
A 30% obstruction and 80% transmission
Obstruction and transmission losses are distinct, and both remain visible in the area ledger.
Aclear=π(D²−d²)/4; Aeffective=T·Aclear; collection ratio=Aeffective/Aref; area obstruction=(d/D)²; Dequivalent=D√[(1−(d/D)²)T].
Supported inputs
Precision and limits
Comparable entered light conditions
The ratios assume the same incident source flux and comparable spectral transmission conditions. Coatings, wavelength response and illumination are not inferred from an instrument type.
Collection is not resolution or visibility
The equivalent diameter and flux ratio do not model diffraction, contrast, eye response, seeing, exposure noise or a limiting stellar magnitude.
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