Some thoughts on "Distance to Station"

In the mist enshrouded past, when dinosaurs roamed the Earth and Ernie Gann was shooting AN range approaches into Pittsburgh in a DC-2, it was pretty important to have a way to estimate the distance from current position to a station with fair accuracy. It helps to know that the miles will run out before the gallons.

Of course, back then there were only occasional ADF beacons and the ever popular broadcast band radio stations to provide the needed signal. So Ernie would dial up KDKA and listen to Jack Benny or The Shadow, and wonder "Hmmm. How far am I from Steeltown?"

Ernie knew about the "Rule of 60", which says something like this: For small angles (say, 10º or less), at a distance of 60 nautical miles from the vertex, each degree of angular offset is approximately equal to 1 nm offset. Got it? Didn't think so. Let's look at a picture:

image001 (1K)
Figure 1

If the distance D (and for that matter D') is 60 nm, then for small angles (say, θ = 10°) the distance d is equal to 1 nm for each degree of θ.

Now, is the "Rule of 60" precisely true? No. Is it pretty close (and likely good enough for a dark and stormy night over the Allegheny Graveyard)? Yup.

If you remember your high-school trigonometry, you'll recognize that sin θ = d/D' and that tan θ = d/D. So d = D' sin θ = D tan θ. Here's a little table that shows just how much error is in the "Rule of 60" for angles up to 10º:

Table I
θ θ (radians) sin θ 60 sin θ % "error"
1 0.017453 0.017452 1.047144 4.714439
2 0.034907 0.034899 2.09397 4.69849
3 0.05236 0.052336 3.140157 4.671912
4 0.069813 0.069756 4.185388 4.634711
5 0.087266 0.087156 5.229345 4.586891
6 0.10472 0.104528 6.271708 4.528463
7 0.122173 0.121869 7.312161 4.459437
8 0.139626 0.139173 8.350386 4.379826
9 0.15708 0.156434 9.386068 4.289643
10 0.174533 0.173648 10.41889 4.188907

This shows that if you "guesstimate" that distance d = 5 miles if the angle θ is 5º, you'll be under by 0.23 miles or about 4.6%. Not bad for Kentucky windage.

The result for the tangent relationship is quite similar:

Table II
θ θ (radians) tan θ 60 tan θ % "error"
1 0.017453 0.017455 1.047304 4.73039
2 0.034907 0.034921 2.095246 4.762308
3 0.05236 0.052408 3.144467 4.815559
4 0.069813 0.069927 4.195609 4.890218
5 0.087266 0.087489 5.24932 4.986396
6 0.10472 0.105104 6.306254 5.104235
7 0.122173 0.122785 7.367074 5.243909
8 0.139626 0.140541 8.43245 5.405626
9 0.15708 0.158384 9.503066 5.589627
10 0.174533 0.176327 10.57962 5.796188

So, back to Ernie in his DC-2 somewhere near Altoona. He's got KDKA dialed in on the ADF and he's heading 270º magnetic. He wants to check his ETA into PIT so he needs to know how far from the station he is right now. He turns 90º North and starts a clock. He's heading 360º now and the relative bearing (RB) to the station is 270º - i.e., it's off his left wingtip. He motors along until the RB is 260° (it's 10° behind the left wingtip) and then he stops the clock and turns west to a heading of 260º, which puts KDKA back on the nose.

Now, it's one of those rare nights, still and smooth. Based on pitch and power, wind and weather, Ernie figures that his groundspeed is 120 knots. He looks down at the timer and sees that the cross-track leg took 3 minutes. Now if he were the co-pilot he'd get out the chart and the paper and the pencil and the whiz wheel and the ephemeris and probably the sextant and calculate that: sin θ = d/D' and therefore D' = d/sin θ but we know that d = (s/60) x t where s = groundspeed in knots and t = time on the cross-track leg in minutes. Now we get: D' = st/60 sin ?

(All together now) Aha! We see from little ol' Table I that 60 sin θ ≈ θ (for small values of θ). And Ernie's the Captain, so he don't need no stinkin' ephemeris. He knows that: D' ≈ st / θ

So he thinks "120 kts times 3 minutes is 360 divided by 10 degrees is 36 miles. I'm doing 2 miles a minute so we'll be over the city in 18 minutes." He turns to the callow and impressionable young co-pilot and casually says "Call Control and tell them we're 20 minutes out."

Incidentally, it makes sense to think about D' because you probably want to know the distance to the station after completing the cross-track maneuver as opposed to the distance back when you started to wonder (which would be D). But if you really want to think in terms of D, you can substitute the tan θ for the sin θ value and be none the worse for wear. The difference is small for small θ's, so Ernie's formula is still D ≈ st / θ.

Also, Ernie could just as well have turned 80° to starboard, setting the station's RB at 280° and allowing it to fall to 270° (the needle always falls, right?) before turning back on course. The trigonometry differs in detail but not in concept, the errors are still in the 4% to 5% range and the "rule of thumb" is still D ≈ st / θ. What if our Intrepid Aviator didn't want to bore a hole in the sky on a non-productive heading for the length of time it took to change the RB by 10°? Is there anything magical about 10°?

Let's look at a number of cases where θ varies but t is constant. For the table below, assume that the time on cross-track is always t = 6 minutes. Obviously, as θ increases the distance to station D decreases. Groundspeed is still s = 120 kts. The second column is D by Ernie's formula (D ≈ st / θ) and the third column is calculated using the "real" formula (D' = st/60 sin θ).

Table III
θ Ernie's D Actual D % "error"
2 360 343.8445 4.69849
4 180 172.027 4.634711
6 120 114.8013 4.528463
8 90 86.22356 4.379826
10 72 69.10525 4.188907

As you can see, the errors are all in a similar (and modest) range. Therefore, we can state a generalized procedure for estimating distance to station: Turn 90° and fly cross-track at a known groundspeed for an arbitrary number of minutes. Note the change in relative bearing to the station (not exceeding 10°). The distance to the station is given by: D = st/θ where D is in nm, s is groundspeed in knots, t is cross-track time in minutes and θ is change in RB in degrees. (Accuracy will be within 5% but will be further degraded by any uncertainty in groundspeed.)

There, that was fun. But in today's environment it may not be as useful to us as it was to Ernie. However, we can use the same concept to develop a formula that may be more useful to us 21st Century Aviators©. Here's the problem: Given a known course and an offset station that is a known distance from our track, determine groundspeed.

image002 (2K)
Figure 2

Reasoning from the concepts already discussed, it's clear that D ≈ st / θ. Hey, no reason we can't solve for groundspeed! Here you go:

s ≈ D θ/t

So let's say I'm on a Victor airway, with the CDI centered on the #1 nav radio and I note another VOR off course to the east. I measure it on the chart (using my precision calibrated knuckle, or maybe even a plotter) and determine that it is offset 25 nm (i.e., that D = 25). I set the OBS for the #2 nav radio to the radial off of that VOR that will indicate it is abeam. (f'rexample, if my magnetic course is 020, I set up the #2 to 290.) Then I just wait for the needle to center. When it does, I start a clock.

I wait until the CDI shows me some deflection that I like - what the heck, let's say 3 dots. That means θ = 6° (2° per dot, right?). At that point I stop the clock and I have my t. Let's say it took a minute and a half. So t = 1.5. All set. I multiply the offset distance times the bearing change and divide by the elapsed time. s ≈ D θ/t ≈ 25 x 6 / 1.5 which means my groundspeed is (lemme see) 100 knots! Now, the actual number (calculated trigonometrically) is: s = D (60 tan θ)/t = 25(60 x 0.1051)/1.5 = 105.1 knots

So, I got a really quick check on groundspeed, accurate within a hair over 5%, without breaking a sweat. Neat. Now you too can amaze your friends and astound your CFI. Have fun.

-- Frank Van Haste

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