A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
From: Frank Reed
Date: 2022 May 6, 05:26 -0700
The standard dip applies to the sea horizon and is well approximated under average refraction conditions by 0.97·sqrt(h) when h is the height of eye in feet, and the result is minutes of arc. Dip short applies when there is an obstruction like an island or a coastline across a bay blocking the sea horizon. Values of dip short are calculated from:
dip_short = X · d + Y · h/d,
where h is height of eye in feet and d is distance to an obstruction in nautical miles. X and Y are forgettable constants (should be X=0.416 and Y=0.566).
Meanwhile the distance to the sea horizon is about 1.16·sqrt(h) with h again in feet and the result in nautical miles (I call that "dip plus a tip" but that only works for folks who know US-style restaurant tipping). If the obstruction is beyond that distance, dip short does not apply, and standard dip is used instead. But how close is standard dip for somewhat nearer obstructions?
I found an interesting way of writing the relationship between standard dip and dip short. Suppose we call the ratio of distance to sea horizon distance R:
R = d / d_horizon
For example, at a height of eye of 25 feet, the distance to the standard sea horizon would be 5.8 nautical miles. If an island blocks the sea horizon and is located 2.9 nautical miles away then R is 0.5. We can also express the dip short as a ratio to the standard dip. I'll call that S:
S = dip_short / dip_horizon.
Note that S is greater than one; if there's an obstruction, the correction we want to apply to an altitude is larger than the standard dip. It turns out that, when you get the factors in the various renditions of the dip short relationship aligned correctly (X and Y above), there's a simple connection between S and R:
S = (R + 1/R) / 2,
dip_short = dip_horizon · (R + 1/R) / 2.
I haven't encountered this relationship before, but I like it. Continuing the example above with d_horizon=5.8nm and an island at d=2.9nm, R is 0.5 so S is (0.5+2)/2 or1.25. And since you can calculate the standard dip for any height of eye in your head relatively easily, you can now calculate the dip short in your head. It's simply 25% greater than the standard dip (for R=0.5). So for a height of eye of 16 feet, where the standard sea horizon dip is 4', the dip short would be 5'. If your height of eye is 100 feet, implying standard dip of 9.7', the dip short would still be 25% greater or 12.2'.
Shorthand rule: if the distance to the obstruction is halfway to the (hidden) sea horizon, the dip short is the standard dip plus 25% for any height of eye.
A more extreme case: if R=0.1, implying that the nearby obstruction is just 10% of the distance to the sea horizon, then S=5. That means that the dip short correction is five times the standard dip for any height of eye. Note that the height of eye is assumed to be "small" relative to the distance to the obstruction in all cases. So if the distance to the horizon is 5.8nm and the distance to the obstruction is 0.58nm (implying R=0.1 and thus S=5), then for the approximation of the dip short equation to hold, the height of eye should be less than about 10% of the distance to the obstruction so less than roughly 350 feet to be safe.