Originally Posted by
Tweetdrvr
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.