## How Many Pixels

Let's compute the number of pixels using the following assumptions and our geometries from the previous page:

inner diameter = 36 mm

outer diameter = 120 mm

radial resolution = 0.1 um

tangential resolution = 0.1 um (worst case)

In the radial direction (let's call this Y) our formula is:
```
Outerradius - Innerradius           60000 um - 18000 um
----------------------------  =  ---------------------    =   420,000 pixels
resolution (um/pixel)                0.1 um/pixel

```

In the tangential direction we have to take our longest path (2π Outerradius) and divide it by our tangential resolution to get the number of pixels.

```
2π Outerradius           2 * (3.14159) * 60000 um
---------------------  =   -------------------------    =  3,769,908 pixels
resolution (um/pixel)            0.1 um/pixel

```

Now, if we multiple these two numbers we'll get the total number of pixels. And its pretty big.

```
4.2E5 * 3.769E6 = 1.58E12 pixels

```

Converting into GB of memory required to hold these pixels gives us:

197.9 GBytes

Note - I've been advised that the tangential direction a slightly larger value of pixel spacing works better. It turns out that the number of optimal pixels is 3,200,000. If one uses this value then the number of pixels = 1.34E12 and the memory required is 168 GBytes.

Oversampling?

When mastering engineers say that this approach leads to oversampling what they mean is that there are more pixels than really needed in the tangential direction for all but the outer cylinder or strip. As we computed earlier, the pixels for the outer cylinder strip are spaced at 0.1 um because there are 3.7 million spread over 377 mm. But for the inner cylinder there are still 3.7 million pixels but they are spread over 113 mm so they are only spaced 0.03 um apart.

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