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I am working through a paper from Inoue et al. 2013.

In section 6.1, pg 18 it states,

... assuming a whole sky rate of ∼ 600 GRBs/year, a 5◦ diameter FoV and a 10% duty cycle, the telescopes will cover a patch of the sky where a GRB is expected to go off only once every ∼ 35 years

where "GRB" refers to gamma ray burst and "FoV" to field of view.

Can anyone provide some guidance on how this calculation was done?

I can see initially taking a 10% fraction of the whole sky rate to account for the duty cycle. This brings us down to 60 GRBs/ year. Now I expect I need to relate the $5^{\circ}$ FoV to the total $4 \pi$ sky area, but cannot seem to do this in a way which obtains the answer.

Any guidance would be greatly appreciated. Thanks

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up vote 8 down vote accepted

Maybe you are missing that there are 41253 square degrees over the whole sky?

So your detector covers 1/2100 of the sky with a 10% duty cycle.

The rate of detection will be $600 \times 0.1/2100 = 0.0286$ per year. Hence a detection every 35 years.

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Perfect - just what I needed! – Tom Mar 6 at 20:58

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