Advice needed for exposure time computation for galaxies

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  • Thread starter AdrianD
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In summary, the exposure time needed for an extended object is the product of the electron count multiplied by the area in arc seconds.
  • #1
AdrianD
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I am trying to compute the exposure time needed for an extended object (galaxy), in python.
I have found the following formulas:
Exposure time app # press the help button at the bottom of the calculator for the formulas used
Exposure time calc

Let's take for example the M33 galaxy. It has a surface mag of 23 mag/arcsec^2 and the dimension in arc minutes 73 x 45 or 4380 x 2700 arc seconds.
From the first link:
  • What I am having trouble is understanding how to choose the 'n' value, respectively the radius value? Should I choose an arbitrary value? Does the radius mean the sample of pixels from the 23 mag faint spiral arms?
From the second link:
  • I don't understand how to calculate the solid angle Omega_i for M33 galaxy. It seems the solid angle is somehow related to the the 'n' value from the first link? The formulas seem to be equivalent
Here is a sample from my code:

exposure time function:
def time(self):
       
         # k1,k2 = flux/photon energy
         # flux in W/m^2/nm
         # photon energy in W*sec
         # filter bandwidth in nm
         # telescope aperture in m^2
         # mag in mag/arcsec^2
         # image scale in arc sec/pixel
        k1 = util.flux(self.targetMagnitude, self.angle, self.typeOfBand, self.pressure, self.temperature)[0] / util.PhotonEnergy(self.typeOfBand)
        k2 = util.flux(self.skyMagnitude, self.angle, self.typeOfBand, self.pressure, self.temperature)[0] / util.PhotonEnergy(self.typeOfBand)
       
        radius = 10 # arc sec
        pixelSurface = self.imageScale**2
        npix= np.pi*(radius**2/pixelSurface)
       
        self.targetElectronsSec = self.QE * k1 * self.filterBandwidth * self.effectiveAperture
        self.skyElectronsSec = self.QE * k2  * self.filterBandwidth * self.effectiveAperture * self.imageScale

       

        #solve the  equation for T
        A = self.targetElectronsSec**2
        B = -self.SNR**2 * (self.targetElectronsSec + npix*self.skyElectronsSec + npix*self.darkCurrent)
        C = -self.SNR**2 * npix * self.readNoise**2
        T = (-B + np.sqrt(B**2 - 4 * A * C)) / (2 * A) #returns seconds
 return {'time':T}
 
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  • #2
AdrianD said:
What I am having trouble is understanding how to choose the 'n' value, respectively the radius value? Should I choose an arbitrary value? Does the radius mean the sample of pixels from the 23 mag faint spiral arms?
Looking at the equation below the calculator, ##n## only appears in the noise terms for the sky, dark current, and readout, so it doesn't appear to have anything to do with the target. But playing around with it seems to show that ##n## only affects the target signal, which is odd. It appears that ##n## counters poor seeing. That is, increasing ##n## so that it is above the seeing will get you the full signal from the target. If ##n## is less than the seeing then you'll get less signal (e-/s). I just don't exactly know what this equation means: ##n=π(\frac{radius}{scale})^2##

AdrianD said:
I don't understand how to calculate the solid angle Omega_i for M33 galaxy. It seems the solid angle is somehow related to the the 'n' value from the first link? The formulas seem to be equivalent
I think the solid angle refers to the solid angle for each pixel (or group of pixels if binning), as it is labeled as: solid angle subtended by the integration element.

I believe an 'integration element' is a pixel or pixel group.
 
  • #3
Drakkith said:
Looking at the equation below the calculator, ##n## only appears in the noise terms for the sky, dark current, and readout, so it doesn't appear to have anything to do with the target. But playing around with it seems to show that ##n## only affects the target signal, which is odd. It appears that ##n## counters poor seeing. That is, increasing ##n## so that it is above the seeing will get you the full signal from the target. If ##n## is less than the seeing then you'll get less signal (e-/s). I just don't exactly know what this equation means: ##n=π(\frac{radius}{scale})^2##I think the solid angle refers to the solid angle for each pixel (or group of pixels if binning), as it is labeled as: solid angle subtended by the integration element.

I believe an 'integration element' is a pixel or pixel group.
Signal - help I did found a help file for an older calculator.
So, for extended objects the electron counts are multiplied by the area of a pixel in arc seconds. For this we use the plate scale, that determines the size of the pixel in arc seconds. If my plate scale is 0.5 arc seconds per pixel, and the CCD pixels are square, we multiply the electron count by image_scale^2.
 
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