Shouldn't they curve to correctly represent the curve of ball earth? I mean, the pilots are navigating a ball, right, not a flat plane:
Physical Observer,
If by “correctly” you mean accurately and by “represent” you mean (proportional to the display size) then:
I did the math and the curve that most PRACTICALLY represents the earth for a static display (the horizon width) of approx.3” (=76.2mm) wide is a straight line. Besides, I seriously doubt that depiction of the horizon in this instrumentation is manufactured to a straight line within error any better than .0001 in/in (see GD&T ASME Y14.5M-1994).
If you disagree please propose a manufacturing method (including metrology equipment to validate) that would more “correctly represent” the curve over a display width of around 3” (=76.2mm). Also, if you think it matters to any practical degree go ahead use any display width from 1/16 of an inch to 100 inches wide (or smaller or larger if you prefer).
I’ll even give you the equation. It can be derived from the center-radius form equation for a circle using x and y coordinates and center coordinates (x=h, y=k) which is:
(x-h)^2 + (y-k)^2= R^2 ( you can look this up almost anywhere and test it out)
Let’s center the circle of interest at x=0, this means h=0
Let k=R, so we have x^2+(y-R)^2=R^2
Let’s solve this equation for y as we are interested in the deviation or vertical distance that the curve of a circle is to a straight line (at y= 0) at a point along the x-axis.
y-0= R-SQRT(R^2-X^2) and y-0=R+SQRT(R^2-X^2) (But we’ll ignore this other solution because we are not looking for the farther y-point on the circle from the straight line).
So we have: y-0=R-SQRT(R^2-X^2) which is not yet convenient so we’ll make some substitutions.
Let y-0= Delta (to represent the difference between the curve and straight line at y=0 the horizon)
Let X=1/2*W where W is the width of the attitude display graphic. This is because we want a line that is centered above the circle which is also centered at x=0 and positive x represents only half of the display width W.
So substituting these terms we arrive at:
Delta (R,W) = R-SQRT (R^2-W^2/4) for W<=2R=Dia. (Diameter = 2 * Radius of a circle)
In other words this says don’t choose a W value larger than the Diameter of the circle or we’ll no longer be looking at a point y on the circle we are interested in.
Where:
Delta=deviation of curve to end point of straight line of width W where W represents the width of the display.
W=width/length of Display (length of horizon on Display) where horizon is depicted.
R=Radius of Earth (or any circle or sphere of size you wish)
Delta (R, W) = R-SQRT(R^2-W^2/4) for W<=2R
A quick informal inspection reveals that with R and W so vastly different in magnitude the answer will be practically 0 but we’ll go through it. Basically for a number as large as R^2 being reduced by a number as small as W^2/4 the equation is practically Delta = R-R= 0
For sake of demonstrating the calculation without involving the importance of considering significant figures (certainty and tolerance of 3959 miles compared to certainty and tolerance of 3 inches) let’s assume that the following values are exact. Pay attention to the number of digits here in these numbers.
R= 3959 mi x (5280 ft/mi) x (12 in/ft) = 250842240 inches
W= 3 inches
Delta = 250842240 inches – SQRT( 62921829368217600 inches – 9 inches/4)= 4.4848 x10^-9 inches = 4.48 nanoinches. Repeat NANOINCHES!
The roundest man made object in the world deviates from a sphere by up to .000003” (3* 10 ^-6 or “3 millionths of an inch”). The roundest man made object has 670 times larger the amount of surface deviation from a sphere than a straight line does to a circle the size of earth over a width or span of 3”. Splitting hairs is not even a figure of speech that would put this level of nonsense in the ballpark.
https://www.newscientist.com/article/dn14229-roundest-objects-in-the-world-created/Is there any criticism more pedantic and impractical than what is otherwise a very reasonable omission of “representative curvature” of earth over 3” of width?
You would have had a much more valid point by criticizing the ground being depicted as brown and featureless or the particular shade of blue used if you are concerned with correct representation.
So what exactly do you mean by “correctly represent” other than to reveal your naive criticism and suggest a most extreme inability/denial to grasp/acknowledge the scale of things?