Triangle corners

[Paul Van Dort]

Hi all, perhaps this has been done before, but for your info, here the entry corners of the triangle based on where you want to put the top (rule book says 45 degrees) . When you make the triangles size 0, you should aim at 120 degrees. If you want the top to be at the zenith, aim at 90 degrees. Here the table with everything in between:

top of triangle
at x degrees
elevation           Size of first and second corner of triangle"   
0                          120.00   degrees
5                          119.89   degrees
10                          119.54   degrees
15                          118.98   degrees
20                          118.19   degrees
25                          117.19   degrees
30                          115.98   degrees
35                          114.57   degrees
40                          112.98   degrees
45                      111.21   degrees
50                          109.28   degrees
55                          107.21   degrees
60                          105.00   degrees
65                          102.68   degrees
70                          100.26   degrees
75                          97.76   degrees
80                          95.21   degrees
85                          92.61   degrees
90                          90.00   degrees

[Motorman]

I don't get this, I thought all equilateral triangles had 120 degree corners regardless of size.
 

[L0U CRANE]

MM, the figure is flown on the surface of a sphere, well, hemisphere...

A way to think of the shapes is as the shortest distances tangent to the cornering radii.

To visualize this - a string to make tangents around the three curves (corrners.) Such lines are not straight: they are curved on a spherical  surface. The turn paths alone can  be considered as on a 'plane' surface - each is a small portion of the flight hemisphere's surface (we don't think of the earth radius curvature on street or road maps, do we?) 

[Trostle]

I don't get this, I thought all equilateral triangles had 120 degree corners regardless of size.

Things change when you fly an equiangular triangle on a hemisphere.  What was explained above is that if an equiangular triangle is infinitesimally small on the surface of a hemisphere, the angles between the legs are all 120^o.   If an equiangular triangle is flown on the surface of a hemisphere, where the base leg extends to be 90o wide (1/4 of the circle), each of the other two legs will traverse a 90^o segment of the hemisphere (where the top of the maneuver will be at the top of the circle) and each of the corners will be 90^o.  Each of the vertical legs will each be like 1/2 of a wingover.

So accordingly, where our pattern requires the top of the triangles to be at 45^o on our hemisphere, the three corners of these triangles will be about 111^o.

Keith

[MikeyPratt]

Keith,
Get them your Stunt-ball or globe and show them, I thought this was over, it’s Ridicules LOL.

Mikey

[M Spencer]

  What was explained above is that if an equiangular triangle is infinitesimally small on the surface of a hemisphere, the angles between the legs are all 120o.

So if we did a 10 foot triangle with five foot corners , we could do a ten foot loop . !        SORRY . 

[Ted Fancher]

OH, dear.  All points on a spherical surface are identical.  Simply fly from one such point at level flight elevation above the ground to a point "halfway" to the tippy top of your hemisphere, down to a third point at level flight elevation utilizing identical corner radii at each "corner" and walk away with a big number.  It really isn't more complicated than that.  I would remind that, from the pilot's location at the center of the spherical airspace provided, all points on the sphere are identical.  Simply "draw" the required segments as though they are on a blackboard from your (the pilot's) perspective.

I would also point out that doing the above with the "inputs" originating from a more or less "fixed" location in the center of the sphere makes the whole concept a lot simpler!

p.s. I might also point out that if the pilot is constantly moving the various parts of his/her anatomy during the inputs required the whole simple picture pretty much is at risk.

Ted

[Motorman]

How does this help me beyond an exercise in geometry. You're saying the first turn into the triangle should look like 111 degrees to me? Or is it actually 111 degrees in science but still looks like 120 to me?

[Dan McEntee]

How does this help me beyond an exercise in geometry. You're saying the first turn into the triangle should look like 111 degrees to me? Or is it actually 111 degrees in science but still looks like 120 to me?

     There is no way in hell that a lower skill class pilot can disseminate what a turn in a maneuver looks like in degrees . Just turn the corner as smoothly as you can and aim the model at the top corner of the triangle at 45 degrees, turn to the down side and aim at the point that you think you should turn at and pull up at something close to the same elevation that you started at. Then just do it again!  If you are lucky, the sides will look even. If not, then adjust the first turn and the second turn accordingly. Work on being consistent at the length of the sides and the height of point and the degrees will take care of them selves   When your scores for the trick start getting into the high 20s and low 30s, then worry about degrees! For now, just fly the plane and shoot for basic size and symmetry! You have to know what the shape will look like before you even get there, then just trace the outline with the airplane.  Most of the judges you will fly in front of don't have protractors in their eyeballs anyway!
  Type at you later,
   Dan McEntee

[Trostle]

How does this help me beyond an exercise in geometry. You're saying the first turn into the triangle should look like 111 degrees to me? Or is it actually 111 degrees in science but still looks like 120 to me?

All of the turns in our equiangular triangles on our hemisphere where the bottom legs are at the bottom of the hemisphere (4' to 6' above the ground level) with the top of the triangles at about 45^o elevation will have the turns to be approximately 111^o, NOT 120^o.  In the real world of flying our equiangular triangles on our hemisphere, there is not much difference between 111^o and 120^o turns.  Judges do not carry a protractor with them to measure these angles.  Even if they did, they would seldom be in a position to accurately measure just one turn/leg of a triangle.  Just another note, the rule book does not mention anything about 120^o turns in the triangles or the hourglass.  In fact, the rule book does not specify any of the sharp turns in the pattern in terms of degrees.

Judges probably cannot differentiate the difference if they see a 120^o or a 111^o turn.  What they can or should see is when the sides of the triangles do not look to be the same length, then they will know that the angles have not been flown properly and the figure is not symmetrical.



Keith

Just make all three sides the same length and flown on great circle paths with the top at 45^o and all corners having a "tight radius" and you will get an excellent score.

[Paul Van Dort]

Thanks for all the feedback.

Very often triangles are flown way too large. If the top corner is at the zenith, than you must go up vertically (90 degrees) to be able to fly a decent triangle. This understanding will avoid making triangles that are too flat, making the last corner very difficult.

Related: A good exercise is to draw the triangle that you want to fly on a piece of paper. Rotate the paper 120 degrees and the triangle should still look the same. 

Enjoy stunt!

[Ted Fancher]

How does this help me beyond an exercise in geometry. You're saying the first turn into the triangle should look like 111 degrees to me? Or is it actually 111 degrees in science but still looks like 120 to me?

HI motorman.  First: No I'm suggesting you forget the angles/degree stuff that are impossible to utilize successfully.  Instead, I encourage you to review the post I've pinned to the top of this forum: (Where/how to look when flying the pattern...etc.)  It addresses my comments immediately below in greater length.

If the pilot is constantly moving the various parts of his/her anatomy during the inputs required the whole simple picture pretty much is at risk.  The best trick I know of is to...WHILE FACING THE CENTER OF THE MANEUVER...make miniature versions of the maneuver in question with your hand as it follows the ship.  Set your neutral with the handle vertical and providing the ability to give the vast majority of inputs to the airplane with lower arm/hand motion modestly accompanied by whatever upper arm motion is necessary to get that lower arm, etc. still pointing at the airplane in the overheads/top of hourglasses etc.  IE, MINIMIZE "EXTRANEOUS" CONTROL INPUT MOTIONS TO THE GREATEST DEGREE POSSIBLE AS YOUR HANDLE FOLLOWS THE SHAPE/DEMANDS OF THE MANEUVER.   

A triangle is easy to draw on a blackboard with just wrist, finger and arm inputs "directed at the target...in this case your stunt ship."  Forget the "degrees" stuff and simply fly a unilateral triangle of the proper elevation requirement...half way to directly overhead...and then do it again.

Again, I ask that you track down that video of Brett Buck flying the pattern (I'm not smart enough to tell you how to find it but others will likely help out).  He flies championship maneuvers of accurate sizes literally doing as I mentioned (and tried to do myself when I competed) as though he was drawing them on a blackboard in front of him...only difference being the blackboard was 60-70 feet away so a long piece of chalk is necessary!!! (That's supposed to be a joke!)

Good luck forgetting the angles and simply drawing the pictures.

Ted

p.s. if any of the above sounded snarky to you it wasn't intended to be so.  Just trying to make what to many seems awfully complex reduced to its basics.


Ted

Final "oh, I forgot!

Simply fly the triangle with EQUAL sides and, BY DEFINITION,  the ANGLES WILL TAKE CARE OF THEMSELVES!!!

REAL "FINAL"  OOPS I FORGOT.  MY GOOD FRIEND KEITH JUST SAID PRETTY MUCH THE SAME SORT OF THING IN A DIFFERENT WAY A COUPLE POSTS EARLIER.  HE'S A SMART GUY I LISTENED TO A LOT OVER THE YEARS BUT DID MY BEST TO PRETEND I PREVAILED WHEN I WOULD ATTEMPT TO REBUT HIM.  IN THIS CASE WE SEEM TO AGREE!

[Tim Wescott]

Setting aside the worry that someone's going to start cutting up poor innocent ping-pong balls again, and maybe building super-light 1/2-A powered things on 70' lines to achieve actual 5' radius turns:

When you're doing your turn angle calculation, remember that -- because of the radii of your corners -- the straights are effectively based on a smaller triangle than one with points at the horizon and exactly at 45 degrees.  So your angles are even a bit more open a bit tighter* than 111 degrees, and dependent on the turn radius.

But, like Dr. Fancher said don't sweat that part.  Instead get a good-looking triangle into your muscle memory, and don't even think of bringing a protractor into the circle.  Do make all the legs the same size, do start the top turn early enough so that the top of the turn kisses 45 degrees (or whatever your chosen top line is if you happen to choose to "cheat" the maneuvers a bit higher).  Don't start milling maneuvers into ping-pong balls so that you can bore people at contests with how well your maneuvers match the rule book.

* Edit.  Oops.

[M Spencer]

Tight Corners , real tight ones , the THING is getting out with no ' bobble ' .
Low Speed , calm air help . But when the pressures on , a nano second
pilot response & eagle eye  is the problem .
 Way I see it .

[Steve Helmick]

I see FAR TOO MANY over-rotating the 1st corner, followed by a 90+ deg. at the top, thus requiring a silly hard corner back to level flight. Getting the first corner right and in the correct position to the wind/judges position is the key. 

The diagram in the rulebook shows the aircraft's turns as being 120 degrees. We're trying to do an equilateral triangle, which by definition has three 60 degree corners, so I never really thought that was a good way to describe it, but it's always been that way. I wonder if perhaps some don't understand that 180 deg. (a straight line) minus 120 deg. = 60 degrees? Not everybody took Geometry in HS (I did...got a decent grade, too).

I see some guys doing a 90 to 120 deg. corner...just a medium to slight change in direction at the tops and it pretty much PO's me. A good set of triangles is a beautiful thing. Howard's are usually top notch. Sometimes Paul's aren't dead downwind, which makes the dives look too steep...or maybe the dives are too steep. Sorry to pick on you, Paul!   

I'm also wondering if some folks problems start with astigmatism in their eyeballs? I hope everybody gets their eyes checked every year. None of this stuff is easy, so don't make it harder!  Steve   

[Dan McEntee]

   In order to know how to crank that first corner, you need to know where the last one will be. You know where it is, you just flew right by it! You judge the length of the bottom side of the triangle from that point. Don't forget to remember where the middle is, because 45 degrees above that is the second corner! This all goes along with what I have always said about learning to use a wide field of vision, and seeing the maneuver as you approach it. If you can "see" it, then all you have to do is trace it. As you practice, look for land marks and such in the back ground to help you. This is all much more easily said than done, but when I have been at my best and have my head on right, this is what I see. Most of us get started with stunt and we have "tunnel vision" and focus on the airplane, and then try to time the trick because we can't see the ground, 45 degrees or 90 degrees over head. It's easy to fall back into also. You don't know how far to crank the turn if you can't see the point that you will end up at and crank the next turn. I think the triangle is possibly the hardest maneuver in the pattern, right there with the hour glass, because they have basically the same elements. They drove me crazy 30 years ago and still drive me crazy today! It's one of the things that I judge myself on how well I am flying and how well the model is flying. If the model won't turn well, you can't make a nice triangle.
   Type at you later,
   DAn McEntee

[Tim Wescott]

I see FAR TOO MANY over-rotating the 1st corner, followed by a 90+ deg. at the top, thus requiring a silly hard corner back to level flight. Getting the first corner right and in the correct position to the wind/judges position is the key....

That seems to be the most common -- or under-rotate the first two, end up with a long inverted second leg followed by a large-diameter turn to level.

I haven't done the exercise in a while (I probably should), but I have an equilateral triangle tacked to my home office wall; I just point my finger at it and follow it through for loops.  This helps get the correct corner angles into my muscle memory, when I'm not distracted by everything else that goes on in an actual flight.

[Larry Renger]

What was described is a figure “D”. I see it all too often. 😢

[Brett Buck]

This is what it is supposed to look like:



Brett

[Dave Hull]

The rulebook appears to be biased in favor of electric planes on this maneuver. Based on Brett's posting, they put a zener on the final leg of the maneuver. How am I supposed to duplicate that with an OS alky burner?  I suppose I can tape a zener to the fuselage and point this out to the judges before the flight and still get full points?

The Divot

[Brett Buck]

The rulebook appears to be biased in favor of electric planes on this maneuver. Based on Brett's posting, they put a zener on the final leg of the maneuver. How am I supposed to duplicate that with an OS alky burner?  I suppose I can tape a zener to the fuselage and point this out to the judges before the flight and still get full points?

  It's 100V, so, I suggest you get one, stand in a puddle of salty water, grab the anode, and stick the cathode into the hot side of the AC line. I think that will resolve your problem.

    Brett

  p.s. seriously, however, in any case, it looks like a zener diode because of the cropping out of the irrelevant details - it's a very simple problem, the picture by itself, without any elaboration or text, tells you what the maneuver should look like. If anyone consistently made the maneuver look like the picture to either themselves or their coach or a judge, they would have it nailed and get 37+ points ever time - because most of them look nothing like that!  I also note that mostly, the corners look - note, I said "look", not measure, since no one is measuring it - much much tighter than this from the real contenders, and that is frequently why they are getting 38+.

    What I think is the important point from Paul's OP is that as the maneuver gets bigger, making the angle wrong progressively distorts the maneuver. What is frequently wrong is that one or more of the angles looks too shallow, making the maneuver too "flat" as noted above. The bigger you make it, the wronger and flatter it gets - and almost everyone makes it (and all other maneuvers) too big. So, the pilot, flying alone, with no coach, tends to fly the maneuver larger and larger because it lets the airplane recover between corners better, they are paying close attention to hitting 120, and then are completely dumbfounded when their score gets killed.
   
   The underlying premise of Ted's point (which I agree with - being, as always, his mindless drone/sock puppet) is that you should probably disregard spherical geometry effects (vs a planar projection, which is what the pilot sees) because they tend to be negligible if the maneuver is the near the right size.

   Put another way, the difference between spherical and planar figures from the judge's position is due to parallax effects, and the larger the maneuver, the bigger the difference. This gives you a double whammy - it's wrong because it is too big, and the shape is also wrong (or you have to pay attention to spherical geometry) because it is too big. You *can* overcome it, but it gets disproportionately more difficult the bigger you make it.

[Lynn Weedman]

OH, dear.  All points on a spherical surface are identical.  Simply fly from one such point at level flight elevation above the ground to a point "halfway" to the tippy top of your hemisphere, down to a third point at level flight elevation utilizing identical corner radii at each "corner" and walk away with a big number.  It really isn't more complicated than that.  I would remind that, from the pilot's location at the center of the spherical airspace provided, all points on the sphere are identical.  Simply "draw" the required segments as though they are on a blackboard from your (the pilot's) perspective.

I would also point out that doing the above with the "inputs" originating from a more or less "fixed" location in the center of the sphere makes the whole concept a lot simpler!

p.s. I might also point out that if the pilot is constantly moving the various parts of his/her anatomy during the inputs required the whole simple picture pretty much is at risk.

Ted

right on Ted