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Judging distance during ground refrence

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Judging distance during ground refrence

Old 01-03-2014, 07:31 AM
  #11  
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From the USAF VFR portion of 11-217
A 60 degree angle equals AGL altitude x 1.7 in NMs
A 45 degree angle equals AGL altitude in NMs
A 30 degree angle equals AGL altitude x 0.6 in NMs

So at 1000' AGL in a C 172, half the strut is about 45 degrees equals about 1nm from the runway.

If the runway was 2/3 up the strut or just a bit below the tie down ring, then the pattern spacing would be TOO WIDE at about 1.7-1.9 nm from the runway.

If the runway was 1/3 of the way up the strut or a bit above the wheel, then you are a bit tight at about 0.6 nm from the runway.

YMMV
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Old 01-03-2014, 08:44 AM
  #12  
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Originally Posted by Tweetdrvr View Post
From the USAF VFR portion of 11-217
A 60 degree angle equals AGL altitude x 1.7 in NMs
A 45 degree angle equals AGL altitude in NMs
A 30 degree angle equals AGL altitude x 0.6 in NMs

So at 1000' AGL in a C 172, half the strut is about 45 degrees equals about 1nm from the runway.

If the runway was 2/3 up the strut or just a bit below the tie down ring, then the pattern spacing would be TOO WIDE at about 1.7-1.9 nm from the runway.

If the runway was 1/3 of the way up the strut or a bit above the wheel, then you are a bit tight at about 0.6 nm from the runway.

YMMV
ive measured it with extended/offset track functions on various gps units. Which seat you are in and L vs R traffic skews things significantly. You can use the strut to judge distance, but its not as constant/consistent without taking these factors into consideration. Fairly consistently the people who claimed they were at 1/2 mile were usually at .75 or so. If I remember correctly about 1/2 up the strut if the runway is on your side, 1/3rd up the visible strut if it's opposite. May not be accurate from my memory! There are a lot of other ways to judge some of this stuff. Timing, known distances like the runway, etc.
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Old 01-03-2014, 10:35 AM
  #13  
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Originally Posted by Tweetdrvr View Post
From the USAF VFR portion of 11-217
A 60 degree angle equals AGL altitude x 1.7 in NMs
A 45 degree angle equals AGL altitude in NMs
A 30 degree angle equals AGL altitude x 0.6 in NMs

So at 1000' AGL in a C 172, half the strut is about 45 degrees equals about 1nm from the runway.

If the runway was 2/3 up the strut or just a bit below the tie down ring, then the pattern spacing would be TOO WIDE at about 1.7-1.9 nm from the runway.

If the runway was 1/3 of the way up the strut or a bit above the wheel, then you are a bit tight at about 0.6 nm from the runway.

YMMV
You have mistaken the 45° equation: if you are 1000' AGL, looking down at a 45° angle to a point, you are horizontally 1000' from that point. You have formed an isosceles right triangle.
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Old 01-03-2014, 10:47 AM
  #14  
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Default Distance = Rate * Time

This is a good time to re-enforce practical applications of the Distance / Rate / Time equations.

You should know the the speed of your plane, and your desired pattern size, you can describe the distance in terms of time.

Example: If you're in a C-172, at a pattern speed of about 90 kts, you're making about 1.5NM per minute (or about 1.7 statue miles per minute). If you want your downwind to be 1/2 mile away, parallel to the runway, that is about 17 seconds away.

Use some rough estimation, that you're still making distance in your turns, then you probably want your crosswind and base legs to be about 15 seconds.

I think you'll find that someone who cannot judge distance accurately can still estimate where they'll be in 15 seconds. Ask the student to count out loud to 15 on their cross/base legs, and decide if they are too close or too far.

Also, talk through the implications if they're going faster or slower; if they have a tailwind/headwind. Get them to actually pencil out the math, or make it an E6B exercise.
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Old 01-03-2014, 11:10 AM
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Originally Posted by EasternATC View Post
You have mistaken the 45° equation: if you are 1000' AGL, looking down at a 45° angle to a point, you are horizontally 1000' from that point. You have formed an isosceles right triangle.
I only typed what I found in the USAF Manual from low level and VFR flying from the paragraphs to estimate distance. It seemed accurate enough. At 30,000 feet stuff that is a 45 degree angle away is certainly about 30 miles from the aircraft. It certainly isn't 6 miles away. But your geometry does seem correct.
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Old 01-03-2014, 11:49 AM
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TLAR method is all you really need
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