PPI worksheet

1. Pixel density: 77.26 PPI

1. Find how many pixels fit along the diagonal, using the right-triangle rule (the Pythagorean theorem)

   $d$$=\sqrt{{\mathrm{1,800}}^2+{\mathrm{1,200}}^2}$$=\mathrm{2,163.330765}\text{\hspace{0.17em}px}$

2. Divide the pixels on the diagonal by the real diagonal in inches. That gives the pixels per inch

   $\mathrm{PPI}$$=\frac{\mathrm{2,163.330765}\text{\hspace{0.17em}px}}{28\text{\hspace{0.17em}in}}$$=77.261813$

3. Divide each side's pixel count by the density to get its real length in inches

   $\frac{\mathrm{1,800}}{77.261813}=23.297408\text{\hspace{0.17em}in},\frac{\mathrm{1,200}}{77.261813}=15.531605\text{\hspace{0.17em}in}$

4. Find the width of one pixel (the dot pitch): one inch, 25.4 mm, divided by the density

   $\frac{25.4}{77.261813}$$=0.328752\text{\hspace{0.17em}mm}$

2. Pixel density: 107.99 PPI

1. Find how many pixels fit along the diagonal, using the right-triangle rule (the Pythagorean theorem)

   $d$$=\sqrt{{\mathrm{1,475}}^2+{900}^2}$$=\mathrm{1,727.89612}\text{\hspace{0.17em}px}$

2. Divide the pixels on the diagonal by the real diagonal in inches. That gives the pixels per inch

   $\mathrm{PPI}$$=\frac{\mathrm{1,727.89612}\text{\hspace{0.17em}px}}{16\text{\hspace{0.17em}in}}$$=107.993507$

3. Divide each side's pixel count by the density to get its real length in inches

   $\frac{\mathrm{1,475}}{107.993507}=13.658228\text{\hspace{0.17em}in},\frac{900}{107.993507}=8.333834\text{\hspace{0.17em}in}$

4. Find the width of one pixel (the dot pitch): one inch, 25.4 mm, divided by the density

   $\frac{25.4}{107.993507}$$=0.235199\text{\hspace{0.17em}mm}$

3. Pixel density: 87.74 PPI

1. Find how many pixels fit along the diagonal, using the right-triangle rule (the Pythagorean theorem)

   $d$$=\sqrt{{\mathrm{2,100}}^2+{\mathrm{1,275}}^2}$$=\mathrm{2,456.750903}\text{\hspace{0.17em}px}$

2. Divide the pixels on the diagonal by the real diagonal in inches. That gives the pixels per inch

   $\mathrm{PPI}$$=\frac{\mathrm{2,456.750903}\text{\hspace{0.17em}px}}{28\text{\hspace{0.17em}in}}$$=87.741104$

3. Divide each side's pixel count by the density to get its real length in inches

   $\frac{\mathrm{2,100}}{87.741104}=23.93405\text{\hspace{0.17em}in},\frac{\mathrm{1,275}}{87.741104}=14.531388\text{\hspace{0.17em}in}$

4. Find the width of one pixel (the dot pitch): one inch, 25.4 mm, divided by the density

   $\frac{25.4}{87.741104}$$=0.289488\text{\hspace{0.17em}mm}$

4. Pixel density: 93.94 PPI

1. Find how many pixels fit along the diagonal, using the right-triangle rule (the Pythagorean theorem)

   $d$$=\sqrt{{\mathrm{1,700}}^2+{800}^2}$$=\mathrm{1,878.829423}\text{\hspace{0.17em}px}$

2. Divide the pixels on the diagonal by the real diagonal in inches. That gives the pixels per inch

   $\mathrm{PPI}$$=\frac{\mathrm{1,878.829423}\text{\hspace{0.17em}px}}{20\text{\hspace{0.17em}in}}$$=93.941471$

3. Divide each side's pixel count by the density to get its real length in inches

   $\frac{\mathrm{1,700}}{93.941471}=18.096374\text{\hspace{0.17em}in},\frac{800}{93.941471}=8.515941\text{\hspace{0.17em}in}$

4. Find the width of one pixel (the dot pitch): one inch, 25.4 mm, divided by the density

   $\frac{25.4}{93.941471}$$=0.270381\text{\hspace{0.17em}mm}$

5. Pixel density: 67.13 PPI

1. Find how many pixels fit along the diagonal, using the right-triangle rule (the Pythagorean theorem)

   $d$$=\sqrt{{\mathrm{2,000}}^2+{\mathrm{1,100}}^2}$$=\mathrm{2,282.542442}\text{\hspace{0.17em}px}$

2. Divide the pixels on the diagonal by the real diagonal in inches. That gives the pixels per inch

   $\mathrm{PPI}$$=\frac{\mathrm{2,282.542442}\text{\hspace{0.17em}px}}{34\text{\hspace{0.17em}in}}$$=67.133601$

3. Divide each side's pixel count by the density to get its real length in inches

   $\frac{\mathrm{2,000}}{67.133601}=29.791341\text{\hspace{0.17em}in},\frac{\mathrm{1,100}}{67.133601}=16.385237\text{\hspace{0.17em}in}$

4. Find the width of one pixel (the dot pitch): one inch, 25.4 mm, divided by the density

   $\frac{25.4}{67.133601}$$=0.37835\text{\hspace{0.17em}mm}$

6. Pixel density: 98.28 PPI

1. Find how many pixels fit along the diagonal, using the right-triangle rule (the Pythagorean theorem)

   $d$$=\sqrt{{\mathrm{2,200}}^2+{\mathrm{1,300}}^2}$$=\mathrm{2,555.386468}\text{\hspace{0.17em}px}$

2. Divide the pixels on the diagonal by the real diagonal in inches. That gives the pixels per inch

   $\mathrm{PPI}$$=\frac{\mathrm{2,555.386468}\text{\hspace{0.17em}px}}{26\text{\hspace{0.17em}in}}$$=98.284095$

3. Divide each side's pixel count by the density to get its real length in inches

   $\frac{\mathrm{2,200}}{98.284095}=22.38409\text{\hspace{0.17em}in},\frac{\mathrm{1,300}}{98.284095}=13.226962\text{\hspace{0.17em}in}$

4. Find the width of one pixel (the dot pitch): one inch, 25.4 mm, divided by the density

   $\frac{25.4}{98.284095}$$=0.258434\text{\hspace{0.17em}mm}$

7. Pixel density: 94.31 PPI

1. Find how many pixels fit along the diagonal, using the right-triangle rule (the Pythagorean theorem)

   $d$$=\sqrt{{\mathrm{2,300}}^2+{850}^2}$$=\mathrm{2,452.039967}\text{\hspace{0.17em}px}$

2. Divide the pixels on the diagonal by the real diagonal in inches. That gives the pixels per inch

   $\mathrm{PPI}$$=\frac{\mathrm{2,452.039967}\text{\hspace{0.17em}px}}{26\text{\hspace{0.17em}in}}$$=94.30923$

3. Divide each side's pixel count by the density to get its real length in inches

   $\frac{\mathrm{2,300}}{94.30923}=24.387857\text{\hspace{0.17em}in},\frac{850}{94.30923}=9.012904\text{\hspace{0.17em}in}$

4. Find the width of one pixel (the dot pitch): one inch, 25.4 mm, divided by the density

   $\frac{25.4}{94.30923}$$=0.269327\text{\hspace{0.17em}mm}$

8. Pixel density: 105.78 PPI

1. Find how many pixels fit along the diagonal, using the right-triangle rule (the Pythagorean theorem)

   $d$$=\sqrt{{\mathrm{2,300}}^2+{\mathrm{1,075}}^2}$$=\mathrm{2,538.823546}\text{\hspace{0.17em}px}$

2. Divide the pixels on the diagonal by the real diagonal in inches. That gives the pixels per inch

   $\mathrm{PPI}$$=\frac{\mathrm{2,538.823546}\text{\hspace{0.17em}px}}{24\text{\hspace{0.17em}in}}$$=105.784314$

3. Divide each side's pixel count by the density to get its real length in inches

   $\frac{\mathrm{2,300}}{105.784314}=21.742354\text{\hspace{0.17em}in},\frac{\mathrm{1,075}}{105.784314}=10.162187\text{\hspace{0.17em}in}$

4. Find the width of one pixel (the dot pitch): one inch, 25.4 mm, divided by the density

   $\frac{25.4}{105.784314}$$=0.240111\text{\hspace{0.17em}mm}$

9. Pixel density: 74.55 PPI

1. Find how many pixels fit along the diagonal, using the right-triangle rule (the Pythagorean theorem)

   $d$$=\sqrt{{\mathrm{1,300}}^2+{\mathrm{1,000}}^2}$$=\mathrm{1,640.121947}\text{\hspace{0.17em}px}$

2. Divide the pixels on the diagonal by the real diagonal in inches. That gives the pixels per inch

   $\mathrm{PPI}$$=\frac{\mathrm{1,640.121947}\text{\hspace{0.17em}px}}{22\text{\hspace{0.17em}in}}$$=74.550998$

3. Divide each side's pixel count by the density to get its real length in inches

   $\frac{\mathrm{1,300}}{74.550998}=17.437728\text{\hspace{0.17em}in},\frac{\mathrm{1,000}}{74.550998}=13.413637\text{\hspace{0.17em}in}$

4. Find the width of one pixel (the dot pitch): one inch, 25.4 mm, divided by the density

   $\frac{25.4}{74.550998}$$=0.340706\text{\hspace{0.17em}mm}$

10. Pixel density: 109.85 PPI

1. Find how many pixels fit along the diagonal, using the right-triangle rule (the Pythagorean theorem)

   $d$$=\sqrt{{\mathrm{2,200}}^2+{\mathrm{1,000}}^2}$$=\mathrm{2,416.609195}\text{\hspace{0.17em}px}$

2. Divide the pixels on the diagonal by the real diagonal in inches. That gives the pixels per inch

   $\mathrm{PPI}$$=\frac{\mathrm{2,416.609195}\text{\hspace{0.17em}px}}{22\text{\hspace{0.17em}in}}$$=109.845872$

3. Divide each side's pixel count by the density to get its real length in inches

   $\frac{\mathrm{2,200}}{109.845872}=20.028063\text{\hspace{0.17em}in},\frac{\mathrm{1,000}}{109.845872}=9.103665\text{\hspace{0.17em}in}$

4. Find the width of one pixel (the dot pitch): one inch, 25.4 mm, divided by the density

   $\frac{25.4}{109.845872}$$=0.231233\text{\hspace{0.17em}mm}$