The thermal inversions here have heights on the order of 50 meters, so a target height of 200 meters works well. That makes it look like a small island in the sea, or a fair-sized hill on land.
The simulations fall into two groups: ordinary mock mirages without ducting; and ducted mock mirages.
Before you look at the detailed simulations for the two different sub-types, a little comparison of ordinary and ducted mock mirages may be useful. I've placed the observer 5 meters above the top of the inversion in each case, although the inversion that produces a duct is a little higher than the one that doesn't.
Bear in mind that the simple mock mirages are produced by a weaker thermal inversion than the ducted ones. That means that the width of the miraged (inverted) zone is smaller for the plain ones. Try to ignore such quantitative differences, and concentrate on the qualitative differences between the two types.
Let's start by comparing the two mock mirages for targets 40 km away, which puts the target well beyond the horizon, and allows the characteristic features of each type to be visible.
At the right is a simulation of an ordinary mock mirage, without ducting. Notice the structure around an altitude of −8′, where a ray is tangent to the bottom of the thermal inversion (shown shaded, as usual). The mock mirage is the little glitch in the image at that altitude. It sticks out like the overhanging lip of a cliff.
The deformation of the erect part of the image between altitudes of about −7.5′ and −4′ is due to the top of the inversion; it's not part of the mock mirage. However, it is a significant difference between an ordinary mock mirage display, and a ducted one. (This deformed zone is discussed in more detail below).
Now let's look at a ducted mock mirage of the same target, at the same distance. That's shown on the left.
While the top third, or a little more, of the target is undistorted in both situations (because it's seen through air with the standard lapse rate of 6.5K/km), the lower parts are different. In the ducted mock mirage, the overhanging lip is a lot more obvious, as we'd expect from the stronger inversion required to produce ducting.
But what's really striking about the ducted display is that the part of the target between the mock mirage (which is roughly the middle third of the image here) and the undistorted top third is completely missing. The inverted image zone terminates at a horizontal line here, and the undistorted top sits directly on that line.
That sharp horizontal boundary is the top edge of the duct. Rays that enter the duct, like the one at an altitude of −4′, are bent down so sharply that a whole section of the target, just above the inversion, is invisible from the observer's position. The top of the inversion, which is also the top of the duct, looks like a solid horizontal surface that blocks the observer's view of this part of the target. (Look at the big gap between the rays at −4′ and −2′ in the ray diagram.)
Strictly speaking, the zone of the target that seems to be missing here is actually present — but so extremely compressed that it can't be discerned. This is a particular example of an effect that's seen whenever the line of sight can be horizontal near the top of a duct: the duct top generally appears as a horizontal line in the refracted imagery.
At the right is an ordinary mock-mirage simulation for an observer at 45 m height, which is 5 m below the top of the inversion. (The rest of my ordinary mock-mirage simulations are for an observer 5 m above the top of the inversion.) I've moved the target out to 55 km to get a better example of the effect.
Notice that the part of the target above the mock mirage — i.e., the top third — appears slightly stooped, but not seriously deformed. (The stooping is shown by the flattening of its vertex: in the actual target, the vertex angle is a right angle; but here, it appears obtuse.) Despite the stooping, the sides are still straight.
The transfer curve also shows this effect: the top end of it is steeper than the relatively normal bottom part. A steep transfer curve corresponds to a stooped image. But the steep part of the curve is straight; so straight lines in the stooped zone appear straight, although slightly compressed vertically.
Here, the whole upper part of the target is seen through the change in lapse rate at the top of the inversion. But that change acts like a negative (minifying) lens, which reduces the height of the image. This is a simple way of explaining the stooping.
Now compare that situation with the view from just 10 m higher up, at 55 m; that's 5 m above the top of the inversion, instead of 5 m below it.
This higher observer sees the image simulated at the left. Here, the observer looks down into the top of the inversion, and then out of it again, when viewing the upper third of the target — e.g., the rays at −4′ and −6′ altitude. (This top third of the target occupies about the top half of its image , above −7.5′.)
Of course, the negative-lens effect is still at work, so this region is stooped. The stooping is considerably greater than before, which is reasonable, as we now look through the change of gradient at the top of the inversion twice . But the stooping is not uniform: the upper part of the image is flattened much more than the part just above the mock mirage.
The reason for the greater stooping higher up is that the line of sight is tangent to the top of the inversion near the top of the target; see the ray whose altitude is −4′, for example. So rays from near the top of the target have grazing incidence at the top of the inversion, while those lower down meet that surface at a bigger angle. Consequently, the path length in the negative-lens region at the top of the inversion is longer for rays near the top of the target. This makes the minifying effect greater near the top of the target, and smaller near the mock mirage. So the straight sides of the upper part of the target become curved lines in its image.
By the way, notice that the mock mirage itself (the inverted zone in the image) looks very similar in these two cases: changing the height of the eye hasn't made much difference in its appearance. Only the part of the image above the inverted zone depends on whether the observer is above or below the upper surface of the inversion layer.
Copyright © 2008 – 2009, 2012, 2022 Andrew T. Young
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