Modern electronics for boats now use multi-function display screens that can be chart plotters, SONAR displays, or show various other data. The physical size of the screen is generally given only by the diagonal distance across the screen. A secondary dimension is the ratio of the screen width or horizontal dimension to the screen height or vertical dimension. This ratio is called the *aspect ratio. *

Because displays are made by the hundreds of millions for television use, consumer electronic devices have always tended to use displays with the aspect ratio used in television. For many decades the displays for television broadcasting in (what we now call) standard definition had a 4:3 or 1.33 aspect ratio. When television broadcasting switched to a new standard—high–definition television—the screen changed and became wider with an aspect ratio of 16:9 or 1.778.

The marine electronics industry for decades used 4:3 aspect ratio screens for displays, but with arrival of new television standards, marine displays changed to a 16:9 aspect ratio. The display size, however, continues to be described mostly by just the diagonal dimension.

Judging display size simply by the diagonal dimension is often done by consumers. A better comparison is made if screen area is compared. If judging two displays of different aspect ratio, screen area is really the only way to compare them. And even when judging displays of the same aspect ratio, comparing screen area is a better indicator of true display size.

If the only screen dimensions given are the diagonal length and the aspect ratio, we will need to develop a formula to find screen area from just those two factors. With a bit of algebra (that I will shown below in the addendum), we can find screen area from just the diagonal and the aspect ratio.

Given a rectangular screen with a diagonal distance D and an aspect ratio x, we can find the screen area from:

Area = D^{2} / [x + (1/x)]

The relationship tells us that for a given diagonal distance (D) the actual screen area decreases as the aspect ratio (x) increases. If we are just interested in two particular aspect ratios (4:3 and 16:9) we can pre-calculate the denominator for those two values:

Area = D^{2} / 2.0833 (for 4:3 aspect ratio)

Area = D^{2} / 2.3402 (for 16:9 aspect ratio)

Pre-computing the aspect ratio factor makes it very easy to compare screen area for different diagonals in the two aspect ratios. For example, let's compare two displays called "5-inch displays" by their diagonal measure. One is the old 4:3 aspect ratio, and the other is a new 16:9 aspect ratio:

DIAGONAL ASPECT AREA
5 4:3 5^{2}/2.0833 = 12.0
5 16:9 5^{2}/2.3402 = 10.68

The new "5-inch" display actually has about 11-percent less screen area than the older 5-inch displays. We were getting more display in the older 5-inch models due to the aspect ratio.

Another common size for new and old displays s 9-inches for the new model and 8-inches for the old model. Which one really has more screen area?

DIAGONAL ASPECT AREA
8 4:3 8^{2}/2.0833 = 30.72
9 16:9 9^{2}/2.3402 = 34.61

In this comparison, the new "9-inch" screen is larger than the old "8-inch" by 12.5-percent.

Let's compare an old 10-inch display with a new 9-inch display:

DIAGONAL ASPECT AREA
10 4:3 10^{2}/2.0833 = 48.00
9 16:9 9^{2}/2.3403 = 34.61

The old display has quite a bit more screen area, about 38-percent more.

If shopping for a new display, you might consider the cost of the larger screen in terms of a cost-per-screen area. Let's look at two models, their screen diagonals, and their MSRP, and find the cost per screen area. We compare the GARMIN GPSMAP 742xs "7-inch display" ($1,000) and 942xs "9-inch display" ($1,300). The only difference between them is the screen size. (Garmin publishes the MSRP and actual screen height and width. I used actual screen dimensions to calculate area instead of the diagonal; the difference is very slight.)

MODEL COST AREA $/inch^{2}
742xs $1000 20.74 $48.21
942xs $1300 34.65 $37.52

I like that comparison: the screen display area is getting less expensive as the display gets bigger. That makes me more inclined to think I am getting a bargin in the model with the larger display But let's look at the incremental cost of the larger display.

The only difference between the two models is the display size. So we can say that the $300 difference in cost is all due to the increase in display area. Let's see how much we paid for the extra display area:

MODEL COST AREA Δ$ Δinch^{2} Δ$/Δinch^{2}
742xs $1000 20.74 -- -- --
942xs $1300 34.65 $300 13.91 $21/inch^{2}

Now you can go shopping for a new multi-function wide-screen display with a better understanding of what you are getting and how much you are paying for the next larger screen size.

Here is the derivation of the formula for area from diagonal length and aspect ratio:

Given a diagonal (D) for a rectangular display with horizontal (H) and vertical (V) sides in an aspect ratio of x = H/V, find H and V, and then Area. From the Pythagorean theorem we know:

(1) D^{2} = H^{2} + V^{2}

From the definition of Aspect Ratio, we know:

(2) H = xV

Substituting (2) for H into (1) we have

(3) D^{2} = x^{2}V^{2} + V^{2}

Factoring gives:

(4) D^{2} = V^{2}(x^{2}+1)

Taking the square root of each side:

(5) D = V[(x^{2}+1)^{0.5}]

Re-arranging to find V:

(6) V = D / (x^{2}+1)^{0.5}

Now we have V in terms of x and D. We also know H in terms of x and V:

(7) H = xV

This means we can now define H as:

(8) H = x[ D / (x^{2}+1)^{0.5}]

Area is the product of H and V:

(9) Area = HV

Now we substitute (8) for H and (5) for V into (9), giving:

(10) Area = x[D/(x^{2}+1)^{0.5}][D/(x^{2}+1)^{0.5}]

Multiplying out the right side, first we evaluate the numerator:

numerator = xDD = xD^{2}

Then we evaluate the denominator:

denominator = [(x^{2}+1)^{0.5}]^{2} or (x^{2}+1)

Now we can finally express area in terms of x and D, giving us:

(11) Area = xD^{2} /(x^{2}+1)

And simplifying by dividing both numerator and denominator by x, we finally have:

(12) Area = D^{2}/[x + (1/x)]

Readers with a question or comment can post them to a thread in the SMALL BOAT ELECTRICAL forum reserved for that purpose.

Copyright © 2018 by James W. Hebert. Unauthorized reproduction prohibited!

Author: James W. Hebert

This article first appeared February 15, 2018.

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