>>
>>For example: Start from the base leg (normal level flight
>>at the base of the hemisphere - a great circle path), make a
>>90-degree turn (inside turn for example)as if a wing over
>>where the model is climbing up on a vertical great circle
>>path to the center of the hemisphere top. Upon reaching the
>>very top of the circle, make another 90-degree turn (another
>>inside turn in this example), dive directly to the ground at
>>a 90 degree angle to the ground (from the pilots vantage
>>point), then recover upright (another inside turn) with
>>another 90-degree turn. The figure has 3 equal length sides
>>and 3 equal 90-dgree corners. This totals only 270-degrees
>>where any triangle in two dimensions will sum to
>>360-degrees.
>>
>>interesting stuff.
>
>
>Well, close but no banana.
>
>Hard as it might be to believe, I think if you place the
>flat cardboard triangle inside your super large triangle
>it'll still fit perfectly, 120/60 degree corners and all. I
>know this sounds ridiculous, but bear with me.
>
>It might be easier to visualize if you think of the 90
>degree wide triangle starting 45 degrees to the pilots left,
>climbing on a great circle to directly overhead (all of a
>sudden the "vertical" climb won't look so vertical), turn so
>that a great circle brings you "straight" down to a point 45
>degrees to the left of the pilot and pull out back at level
>flight.
>
>Again, from the pilot's percpective, this triangle will
>appear to be equilateral with three equal sides and three
>equal angles that sure as heck look like 120 degree corners
>to the pilot!
>
>Interesting to note that this is precisely the same maneuver
>that will look totally distorted to the pilot if he were to
>fly the first climbing leg directly in front of him. It
>will appear to have a "too" vertical climbing leg and a
>"too" inverted or flat descending leg. Yet, if he faces it
>directly, it will look like a perfect equilateral triangle.
>do do do do, do do do do!
>
>Point of view is everything! That's one of the reasons it
>is "SO" important for the pilot to face the center of his
>maneuvers! If he flies them off to the side he will have to
>distort the shape just to make it look right to him. Let's
>not even go there!
>
>The huge triangle is a great way to realize the visual
>distortion to the pilot. Thankfully, as the maneuver gets
>smaller the distortion is diminished.
>
snip
>
>Ted
Hi Ted,
This is almost as much fun as the "discussion" we had at VSC or whereever it was a year or so ago.
I think the triangle you describe above is essentially the same triangle I described. We both described triangles where each leg is a great circle route over a 90-degree segment of the hemisphere. To the pilot, this will appear as an equiangular triangle where indeed a flat plate can be inserted in the cone with a triangular cross section. The pilot sees 120-degree corners on that flat plate, yet the airplane is only turning 90-degree corners on the three separate planes that are tangent to the hemisphere at each of the three corners.
The same thing basically happens with the square loops where all four sides are of equal length, are great circle paths and all four corners are the same. The flat plate that describes the cross section of this four-sided cone is a true square. The pilot sees a 90-degree turn at each corner. But the angles the airplane is actually turning in the planes tangent to the hemisphere at those corners are each less than 90-degrees. The airplane is moving on the three dimensional surface of a hemisphere. It is not moving in the two dimensional flat plane that describes the cross section of the cones we are talking about.
From your post #48, you stated:
"Try to get your mind around this one. You could fly a "square 'overhead' loop" of 179 degrees included angle whose corners would amount to little more than a tiny climb every ninety degrees around the circle. You could still cut that flat square and fit it inside the maneuver "cone" just as with the smaller sized ones."
Ted, you are correct and I think with this example as well as the triangle you described above, you just helped me make my case. The square maneuver you describe would be what is projected from that flat square to the surface of the hemisphere and would still appear as a square maneuver to the pilot. However, the turn angles on the surface of the hemisphere are far less than 90-degrees. (It is just that the pilot is almost in the same plane as the flat plane that contains each of the four corners of the square rather than looking at that plate from a distance as he is doing with the "45-degree square loops" of the rulebook.)
Please be prepared to be enlightened when you get to Tucson for VSC. The cones I have and the triangles that I use to show this on my clear hemisphere are very accurate and quite enlightening.
)