1. **Stating the problem:** We have LED points located at specific coordinates in inches on a 200" diagonal image. The display resolution is 1024 x 768 pixels. We want to find the corresponding screen coordinates for these points on the LCD screen. 2. **Understanding the problem:** The physical size of the image is 200" diagonal. The resolution is 1024 pixels wide and 768 pixels tall. We need to convert the physical coordinates (in inches) to pixel coordinates. 3. **Calculate the physical width and height of the image:** The aspect ratio is 1024:768, which simplifies to 4:3. Let width = $W$ inches and height = $H$ inches. Using Pythagoras theorem for diagonal: $$W^2 + H^2 = 200^2$$ Since $\frac{W}{H} = \frac{4}{3}$, we have $W = \frac{4}{3}H$. Substitute: $$\left(\frac{4}{3}H\right)^2 + H^2 = 200^2$$ $$\frac{16}{9}H^2 + H^2 = 40000$$ $$\frac{16}{9}H^2 + \frac{9}{9}H^2 = 40000$$ $$\frac{25}{9}H^2 = 40000$$ $$H^2 = 40000 \times \frac{9}{25} = 14400$$ $$H = \sqrt{14400} = 120$$ Then: $$W = \frac{4}{3} \times 120 = 160$$ So the physical dimensions are 160" width and 120" height. 4. **Calculate pixels per inch (PPI):** $$PPI_x = \frac{1024}{160} = 6.4$$ $$PPI_y = \frac{768}{120} = 6.4$$ 5. **Convert each LED coordinate (in inches) to screen coordinates (pixels):** The origin (0,0) in inches corresponds to the center of the screen. Screen coordinates: $$x_{screen} = (x_{inch} + \frac{W}{2}) \times PPI_x$$ $$y_{screen} = (\frac{H}{2} - y_{inch}) \times PPI_y$$ Note: y is inverted because screen y increases downward. 6. **Calculate for each point:** - (-50, 50): $$x = (-50 + 80) \times 6.4 = 30 \times 6.4 = 192$$ $$y = (60 - 50) \times 6.4 = 10 \times 6.4 = 64$$ - (0, 50): $$x = (0 + 80) \times 6.4 = 80 \times 6.4 = 512$$ $$y = (60 - 50) \times 6.4 = 10 \times 6.4 = 64$$ - (50, 50): $$x = (50 + 80) \times 6.4 = 130 \times 6.4 = 832$$ $$y = (60 - 50) \times 6.4 = 10 \times 6.4 = 64$$ - (-50, 0): $$x = (-50 + 80) \times 6.4 = 30 \times 6.4 = 192$$ $$y = (60 - 0) \times 6.4 = 60 \times 6.4 = 384$$ - (50, 0): $$x = (50 + 80) \times 6.4 = 130 \times 6.4 = 832$$ $$y = (60 - 0) \times 6.4 = 60 \times 6.4 = 384$$ - (0, -50): $$x = (0 + 80) \times 6.4 = 80 \times 6.4 = 512$$ $$y = (60 - (-50)) \times 6.4 = 110 \times 6.4 = 704$$ - (-50, -50): $$x = (-50 + 80) \times 6.4 = 30 \times 6.4 = 192$$ $$y = (60 - (-50)) \times 6.4 = 110 \times 6.4 = 704$$ - (50, -50): $$x = (50 + 80) \times 6.4 = 130 \times 6.4 = 832$$ $$y = (60 - (-50)) \times 6.4 = 110 \times 6.4 = 704$$ **Final screen coordinates (pixels):** - (-50, 50) → (192, 64) - (0, 50) → (512, 64) - (50, 50) → (832, 64) - (-50, 0) → (192, 384) - (50, 0) → (832, 384) - (0, -50) → (512, 704) - (-50, -50) → (192, 704) - (50, -50) → (832, 704)