A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
From: Brad Morris
Date: 2013 Mar 10, 09:24 -0400
When I first considered the wave height correction, the distance to the wave was
sqrt(h^2 + (3860*sqrt(h))^2)
because we are sighting along the diagonal of a right triangle. Lets examine that with Jeremy's height of eye (h) = 103 feet = 31.312 meters.
sqrt(31.312^2 + (3860*sqrt(31.312))^2)
Now lets do it by the proposed distance to the horizon
When considering the foreshortening of the wave height by the viewed angle, I considered
since we are at elevation and viewing it at a downwards angle, the dip angle. At h=31.312 meters, the dip corr'n from the NA is 9'.8. cos(9.8/60) = .9999995 Lets assume a 30 foot wave! So its perceived to be 4.5599 meters tall at Jeremy's bridge instead of 4.56 meters (a 100 micron difference).
So instead of (cos(dip)*wvht/2) / sqrt(h^2 + (3860*sqrt(h))^2), I simplified it to
When I want to find the angle in a right triangle, I use the arctangent function. You may use whatever you prefer. The arctangent button on my calculator works just fine, thank you! I don't like extra rules or extra steps to avoid a trig function, as you proposed. But hey, thats just me.
So for the example of 30 foot seas from Jeremy's bridge we have
atan(4.56 / 21599.4)
=.012096 degrees = 43.54 seconds
You proposed subtracting the wave height directly from the height of eye (something I did consider) and then determining dip.
So for the example, h =31.312 meters and wave height / 2 = 4.56 meters. Your effective height of eye is 26.752 meters. Interpolating in the NA dip table yields a dip corr'n of 9'.1128.
This is a wave height correction of 41.23 seconds.
For any practical purpose, not just when wvht is 'small', the equation derived is functional and efficient.
Yup, simple trig!
Brad, you proposed this equation for calculating the impact of wave height on dip:
"The angular height of waves at your horizon is
wvhtCorr'n = atan((wvht/2)/(3860*sqrt(h)))*60
where wvhtCorr'n is the waveheight correction, in minutes
wvht is the measure of the larger of wind waves or swell at the buoy
which is a peak to trough measurement, we want mean sea level
to peak. Meters
h height of eye, meters
You may recognize the term 3860*sqrt(h) as this is the distance to the visible horizon. So we are taking the arc tangent of Y over X, which is 1/2 the peak to trough wave height divided by the distance to that wave height! Simple Trig!"
No, not simple trig....
First, you don't need the arctangent here. Nor do you need an arcsine or any other "simple trig". The angular size of anything, when the angle is very small, is given by its physical size divided by its distance. Need to know the angular height of a lighthouse in the distance? Take its height, divide by distance, and convert to minutes of arc (see below). If you need to, you can justify this simple ratio from a "trig" point of view by replacing tan(x) or sin(x) by x which is very accurate for small angles. But more fundamentally, an angle IS a ratio: an angle is the ratio of arc to radius, and for small angles, the length of the arc across the line of sight is nearly indistinguishable from the linear size of the object. Let's call the wave height from sea level to peak, h'. The angular size of that height from some distance, d, is x=h'/d. That's all --just a ratio. Of course to convert it our common angular units, you need to multiply by 57.30 to convert to degrees, or (the number I prefer to remember for navigational work) 3438 to convert to minutes of arc. But save that conversion for last. The equation is just h'/d.
Second, there's a bigger problem. This "simple trig" analysis would work if the waves with height h' were like a wall out there at some specific distance, the same for all observers. But it's not like that. If you are 15 feet above sea level looking out at an ocean with five foot waves (calm sea level to wave peak, not trough to peak) and I am standing on a ladder next to you 25 feet above see level looking out at the same sea, we are NOT looking at the same crest of waves at the horizon. You've correctly accounted for the greater distance but not the fact that they are in a different orientation at that distance due to the curvature of the Earth. Fortunately, there is a much easier way to do this.
Now let's reduce the equation you've been looking at up to this point with a little algebra. Sticking with unrefracted equations for the moment, your dip corrected for wave height is:
dip = dip0 - h'/d
where dip0 = sqrt(2h/R).
But since the dip in minutes of arc and the distance, d, in nautical miles are equal in the unrefracted case, we know that d = R*dip0. If we do a little algebra, your dip equation reduces to
dip = dip0*(1-h'/(2h)).
Nice and simple. But is it correct? The answer is, yes, but only when the wave height is small compared to the uncorrected height of eye. There's a much easier way to picture and calculate the correction for wave height: simply measure YOUR height from the surface defined by the tops of the waves. This is easy enough from a vessel at sea. You look over the side and see where the waves are peaking relative to your position on deck. Of course, if you're on a small vessel, you have the added complexity of waiting for the vessel itself to be near the tops of the waves if they're large enough. If you're an observer on shore, you have to remember that the waves are bigger "out there". But all you're doing then to correct for wave height is taking your height above the mean water level and subtracting the height of the wave tops, h'. The correct dip equation, in other words is:
dip = sqrt(2(h-h')/R).
If you have a dip table, there's nothing else required. You just enter it with the height h-h'. The tables are fine as they are. But let's go a little further algebraically. This correct equation can be re-written:
dip = dip0*sqrt(1-h'/h).
This, too, is an exact equation, but in cases where h'/h is small (normally less than 10-20% is considered reasonable for this), we can replace the square root by the binomial expansion: sqrt(1-x)=1-x/2. Doing that here gives the approximate dip for cases when the wave height is small:
dip = dip0*(1-h'/(2h)),
which, you'll notice, is exactly the same as the equation that you had come up with previously. So now we know that the approach you devised does work under some circumstances, specifically when h' is a small fraction of h. But in general, you should just be using the simple replacement: h-h' --your height of eye should be measured from the tops of the waves, not from sea level. It's easy to forget this when you're standing at a shore location where the waves are much lower in amplitude than they are a few miles offshore. But that's all there is to it.
In the above, I assumed the unrefracted case for dip. The correction for dip in cases of a linear change in temperature with height is just a small proportional change in the "effective" radius of the Earth, as we've discussed before. And I agree with Marcel that this is not, in general, going to be something calculable from measurable parameters, but that's a topic for another message...
PS: There's one more special case of geometry that you can try to puzzle out for fun. What happens if my height of eye is below the wave height miles out to sea (h<h'), but the wave height increases linearly, steadily between my location and some point beyond the horizon? :) In practice, you can avoid this problem, but it's an interesting intellectual exercise.
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