A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
From: Brad Morris
Date: 2013 Mar 13, 15:12 -0400
The entire purpose of the equivalent solutions was to show that they were equivalent. Indeed, they are!
Perhaps my solution as derived is awkward for some. It was independently derived. That is, I illustrated the problem on a piece of paper and then attempted to solve for the correction. It appears that I did, albeit in a different form than you recommend. I don't understand why this is so important.
I also used large height of eye (h=Jeremy's Bridge) to show the solution without exception in this regard vs smaller h. Ditto the large (30 foot) waves. Neither of these would be considered small in the real world.
In simple fact, the largest set of waves I've got on video is of waves breaking on the outside boulder at Half Moon Bay. That's a set of waves at 100+ feet to the crest from the trough. The ocean floor there is mainly 90 feet deep, but the boulder is at 60 foot deep. Thats what this set broke on. Fantastic but true, I have the evidence. The main reef is exposed at low tide and roughly 3-10 feet deep at high tide, yielding some spectacular surfing rides. We call it "Mavericks".
So for the example then, this is roughly a 9 arc-minute correction for wave height. Unless your bridge is higher than this, the dip correction becomes additive, a rather unusual outcome.
In this odd case then, solving for dip as a function of h-h', is the dip of a negative number. The tabular function in the NA does not support this.
Sorry for the delay in replying to this.
Your last post on the wave height calculation started off with this:
"Lets examine that with Jeremy's height of eye [...] =21599.4 meters.
Now lets do it by the proposed distance to the horizon
[...] =21599.4 meters"
Yes, they are nearly indistinguishable. I don't understand why you brought this up. How is this related to my previous comments?? BTW, of course, they are nearly identical distances --this is just the stretched string problem again! :)
And you wrote:
"When considering the foreshortening of the wave height by the viewed angle"
I also don't understand why you're bringing up 'foreshortening'. Foreshortening is irrelevant here. Any object at the horizon that is locally vertical is NECESSARILY perpendicular to the observer's line of sight. There is no foreshortening even in principle.
Brad, you wrote:
"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."
I understand where you're coming from, Brad. Many people who know trig have become comfortable using the atan or the asin button on a scientific calculator as a "quick" means of converting a small angle to degrees. And there's nothing wrong with that, just so long as you realize that this is all you're doing. You're not "solving" a trigonometry problem that actually requires that function. And even in this very specific case, the savings in effort is small to nil. Consider a light house 75 feet tall that's about 15,000 feet offshore. What's its angular height? That's a ratio of 1/200 or 0.005. So what's the angle in minutes of arc? How many keys do you have to press to get the answer by the atan method? Or, since there's really no well-defined triangle here despite the illusion of "simple trig", you could equally well use the arcsine. How many keystrokes then? Or you could skip the trig buttons, as I've suggested, and go directly from the ratio to minutes of arc, by multiplying the ratio by 3438. How many keystrokes then?? Seriously: count them out. Of course, on almost any other computing platform, the inverse trig functions normally return the result as a fraction (or "in radians") so you have to do the multiplication by 3438 (or equivalent) anyway. Next, and this is the REALLY important case, suppose you want to solve this problem in your head. The solution is "divide height by distance and multiply by 3438". It couldn't be easier, and you do not require a calculator or tables. Almost anyone can get an answer accurate to within 5 or 10% without even putting a pen to paper. And note that the last step, the multiplication by that "magic number" (which for perfect accuracy is really 10800/pi) is ONLY necessary if you need to have the result in minutes of arc. Angles are numbers --pure numbers. They have no units. Multiplying by 3438 to "convert" to minutes of arc is like mutiplying 0.75 by 100 to "convert" it to a percentage. We do it because we are "used to" numbers in sexagesimal notation.
Now let's set aside practical calculation (again) and recognize that an equally useful reason for replacing the atan by the pure ratio is algebraic simplification. Your formula for dip corrected for wave height was (slimmed down),
dip = dip0 - atan(h'/d)
where dip0 is the normal dip, h' is the height of the wave (half the height from trough to peak) and d is the distance to the horizon. Now if you drop the atan, or, which is exactly the same thing, replace it by its series expansion and drop very small terms, you get
dip = dip0 - h'/d
But d in nautical miles is equal to dip0 in minutes (no refraction yet). Or if dip0 is a pure angle, d=R*dip0 (R is the radius of the Earth) and of course dip0=sqrt(2h/R). If you work the algebra, you can show that this reduces (exactly) to
dip = dip0*(1-h'/(2h)).
The point here is that this is identical to your original equation as long as h'/d is a small angle, which it always would be in any conceivable case. So we've gone from an equation that requires an atan lookup to something that you can work out in your head in a few seconds. That's the HUGE advantage of recognizing that the "atan" is un-necessary. Let's do an example "in our heads". I'm 25 feet above sea level on the shore. That means that dip0 is very close to 5.0 minutes of arc. I have a source that tells me that typical wave heights a few miles out are around 10 feet from trough to peak. I divide that by two to get h'=5 feet (this is the height of the peaks above mean sea level). Next h'/(2h) is just 5/(2*25) or 0.1 so the result then is that dip = dip0*0.9 or 4.5'. So the dip is reduced by 10% or half a minute of arc by the height of the waves.
As I noted previously, your equation is a good approximation when the wave heights are small compared to height of eye. That's because the CORRECT equation for dip corrected for wave height is very simple: you just replace h by h-h'. This is algebraically equivalent to
dip = dip0*sqrt(1-h'/h)
which, as I described previously, in the case where h'/h is relatively small (and "relatively" here means less than about 0.2), is nearly equal to the reduced version of YOUR equation (since sqrt(1-x)=(1-0.5*x) when x is small). But note that your equation would give misleading results when the wave height (level to peak) is more nearly equal to the observer's height of eye above sea level. You should not use your equation except in cases where the observer's height of eye is significantly larger than the wave height.
Finally, just to reiterate, if you (or anyone!) want to correct for wave height when using a standard dip table, the procedure is very simple: enter the table with your height of eye, h, above sea level (or some equivalent, like your vessel's waterline) minus the height, h', of the peaks of the waves above sea level. Whether you can actually determine these numbers is a whole 'nother can of worms. This type of uncertainty is just one of the reasons why the correct real-world value of the dip constitutes the PRINCIPAL uncertainty in standard line of position celestial navigation.
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