How to fly DME arcs?

[CA1900]

And as an aid to maintaining the arc once you're on it, make a note of the DME groundspeed reading. The closer the groundspeed reads to zero, the less your distance is changing from the VOR. If the mileage is correct and the groundspeed reads zero, you're flying a perfect arc. Neat, eh?

 

[Flyin Tony]

I got a GPS that does all that carp for me www.Echoflight.com

 

[Pilot_Ryan]

What is the best procedure to fly DME arcs when being vectored? How far do you know you need to turn? I have always hear turn 90 degree, but what if the controller vectors you higher along the arc and at an angle?

Based on your profile I'm guessing your Cessna 172 and Arrow aren't equipped with FMS, EFIS, or an RMI...

Calculating lead intercepts for DME arcs and any selected course



TAS / 200

Example:

120KTAS / 200 = 0.6

Therefore, initiate a standard rate turn 0.6nm prior to arriving at the arc.


For a 45 degree intercept to the final approach course, 0.3nm
(use GPS for XTRK information.)


For a 30 degree intercept, initiate at 0.2nm.


This assumes a perfectly timed intercept (doesn't account for wind, roll rate, etc.) For best results I suggest adding .1 or .2 so that you can fine-tune the roll-out as you come close to intercept.

 

[HMR]

TAS / 200

Example:

120KTAS / 200 = 0.6

Therefore, initiate a standard rate turn 0.6nm prior to arriving at the arc.

It has nothing to do with TAS. The turn needs to be predicated off GROUNDSPEED. A proper formula would be GS/100 (not 200).

Example: At a groundspeed of 120kts start the turn 1.2nm from intercept. At 480kts, start the turn at 4.8nm. Try it. You'll like it.

Don't believe me? Do the math: How long does it take to do a standard rate, 90 degree turn (including lag time to initiate the turn)? How far does an aircraft with a GS of 120kts travel in that time?

FWIW- A 90 degree intercept seems to be the most common entry for DME arcs.

 

[bubbers]

Kingston has a DME arc approach that is the standard arrival from the north. At 250 knots at around 6,000 feet initiating the turn 2.5 NM prior to arc worked perfect every time then letting that 727 stay in a 10 degree bank and playing the RMI like a U Control airplane on two wires made it as simple as a straight in approach. Just keep the wingtip on the vor and adjust the DME to maintain the arc. After age 60 we aren't trusted to do that any more.

 

[Pilot_Ryan]

It has nothing to do with TAS. The turn needs to be predicated off GROUNDSPEED.

No, it's TAS.

The airplane drifts with the wind during the turn. The groundspeed is not constant because the heading is constantly changing. Say for example that an airplane flying 120 KTAS has a direct tailwind of ten knots. As the standard rate turn begins to intercept the arc, the airplane drifts by windspeed/120, or about 0.083nm per minute. Round that up to 0.1nm and that's about how far you'll drift during the intercept - tack it on if you're so inclined (the rule of thumb doesn't include this because it's not really worth the trouble.)

In reality, this doesn't matter much. Use groundspeed or TAS and you'll be "about" right. But the correct formula does call for TAS, not GS.

A proper formula would be GS/100 (not 200).

I disagree. The formula I've quoted works in all aircraft I've ever flown, and I've seen it in print (Mental Math for Pilots, IFR Magazine) more than once. Additionally, the concept appears in other forms. Transport Canada's Flight Training Manual advocates a DME arc intercept technique using 0.5% of GS (or TAS) for a standard rate (they call it 'Rate One') turn and 1.0% for a 1/2 standard rate turn. Since the Chickenhawk isn't likely to be going fast enough to ever require a half-standard rate turn for anything, 0.5% of 120 knots = 0.6nm (exactly the same result as the formula I published above.) An airplane with a groundspeed of 200 knots would require 1nm (0.5% of 200 = 1nm) for the lead-in, etc.

Another common rule of thumb is to use a 0.5nm lead-in for groundspeeds below 150 knots. It's not quite as precise - you'll overshoot a bit at the higher speeds - but it essentially works, and is pretty much in line with the formula I provided. As indicated above, add .1 or .2 to account for the roll-in. Add more if you're above 200 knots, but most of us aren't intercepting DME arcs much above that speed, whether we're flying jets or pistons.

In summary, your suggestion of TAS (or GS) / 100 would be accurate for half-rate turns, not standard rate. The piston driver would grossly undershoot, every time.

Don't believe me? Do the math: How long does it take to do a standard rate, 90 degree turn (including lag time to initiate the turn)? How far does an aircraft with a GS of 120kts travel in that time?

The math isn't as straightforward as you seem to think, and doesn't support your inferred conclusion. Take two airplanes flying 120 knots straight ahead; one begins a standard rate, 90 degree turn while the other continues. Start a stopwatch. The speed of the turning airplane relative to the level-wing airplane drops from identical (120 knots) at the beginning of the turn to zero at the thirty second mark. Had he continued straight ahead for one minute, he'd travel about a mile. But since he made the standard rate turn, he only covers about half that relative distance - 0.5nm. I.e., median average of 120 knots at the 0 second mark and 0 knots at the 30 second mark = 60 knots, which would cover 0.5nm.

FWIW- A 90 degree intercept seems to be the most common entry for DME arcs.

Right. But I'd suggest 80 degrees when intercepting from the outside, which is how most of us use DME arcs, most of the time.

Best regards,

-Ryan

 

[fokkers&beer]

It aint rocket science. I typically shoot 2 arcs a day in SE Alaska. In the Fokker I lead it by about 2 miles. How did I come up with this?, elementary my friend, I shot a butload of arcs. In the ATR however because of the slow reaction and Autopilot, I lead three miles. It just takes a few arcs to figure out how fast the aircraft will roll 90 degrees. There is probably a formula, and I will let you UND guys explain how it works.

I myself use 2 or 3 miles depending on the AC.

 

[prpjt]

No, it's TAS.

Since the Chickenhawk isn't likely to be going fast enough to ever require a half-standard rate turn for anything, 0.5% of 120 knots = 0.6nm (exactly the same result as the formula I published above.) An airplane with a groundspeed of 200 knots would require 1nm (0.5% of 200 = 1nm) for the lead-in, etc.

-Ryan

Using the Chicken hawk example, lets approach this a different way. Let's say you have it pulled back to 80 knots TAS. Here we go.

Plane #1 approaches the arc with a 70 knot tailwind. Your method has you starting the turn 0.4 nm out. Your aircraft is travelling 150 knots toward the intercept point. I contend that you'll blow through it.

Plane #2 approaches the arc with a 70 knot headwind. Your method has you starting the turn 0.4 nm out. Your aircraft is approaching the interrcept point at 10 knots. Forget starting a turn at .4 nm, You would have to fly for almost 2.5 minutes straight ahead just to get to the arc.

It's GS.

 

[A Squared]

It's GS.

Yeah, it's groundspeed, obviously, as your example shows pretty conclusively. I guess Pilot_ryan has never really though it through all the way.

 

[HMR]

Yeah, it's groundspeed, obviously, as your example shows pretty conclusively. I guess Pilot_ryan has never really though it through all the way.

Oh, C'mon guys... when do we ever see 70kt winds?

Best Regards,

-HMR