Triangles...how to fix yours

[Steve Helmick]

Hey, you in the back row! Wakeup!!!  Steve

PS: Some might want to start at about the 17 minute barrier. I think triangles are cool!

[Trostle]

Interesting and informative.

However, the triangles flown in our stunt pattern are on a hemisphere, have somewhat round corners and the sides are to be great circle paths on the surface of that hemisphere.  The way the triangles are described in the rulebook is that the sides are to be of equal length and the corners are equal size.  Thus, our triangles are to be equilateral in shape (with sort of rounded corners).  An interesting thing happens when you put an equilateral triangle on the surface of a hemisphere.  The sum of the angles between the sides is NOT 180^o.  The larger that triangle is with respect to the size of the hemisphere, the sum of those three angles becomes more and more.   This means that to properly fly our equilateral triangles with the top at the 45^o elevation, our corners will have to be less than 120^o.

It would be interesting to hear a similar lecture on equilateral triangles in spherical geometry.  There are not many text books that discuss this.

Keith

[Serge_Krauss]

Keith -

I think you have a typo to correct. The sum becomes greater (I think the limit is probably 360^o).

SK

[Jim Thomerson]

Sum of the angles of a spherical triangle range from 180 to 540 degrees.  A point triangle would sum 180; a triangle expanded to become a great circle would have 540.  However. the pilot sees the triangle as if it were on a flat surface, so can enjoy the illusion that it is a 60 degree corner equilateral.

[Tim Wescott]

Keith -

I think you have a typo to correct. The sum becomes greater (I think the limit is probably 360^o).

SK

Indeed.  I started this to contradict you, and realized you were right.  Keith is correct that the amount you turn is less, but that is because the amount you turn is the compliment of the angle that you measure on the triangle.

Visualize this maneuver: do a right-angle turn, as if to start a wingover.  At the peak of the circle, do another right angle turn.  Now do a right angle turn again, to recover level at 5 feet.  Tra-la -- you have just done an equilateral triangle, with three 90 degree corners -- that adds up to 270 degrees.

Where Keith is right is that on a flat surface, the equilateral triangle requires you to turn thrice at 120 degrees, for a total of 360; in my 90-degree example you must turn thrice at 90 degrees, or only 270 degrees.

For the degenerate case (put away those whips and the leather underwear -- I mean mathematically degenerate) fly a level lap.  When you pass your flight box say "turn", but stay level.  Now fly 1/3 of the way around the circle and say "turn" again, but fly level.  Now "turn" again at the 2/3 point.  Congratulations!  You have just inscribed the largest equalateral triangle that you can on the surface of a sphere.  All the angles were 180 degrees, which meant that you only had to turn zero degrees.

[Dan Bregar]

Wow, really good information !  However I'm having trouble operating my calculator while flying the pattern. I must be doing something wrong. 

[Trostle]

Keith -

I think you have a typo to correct. The sum becomes greater (I think the limit is probably 360^o).

SK

Serge,

You are right.  The way I worded it is that the sum of the angles between the sides do become more and more when you fly larger and larger equilateral triangles on the surface of a hemisphere.  The actual angles turned do become less and less.  So, the angles between the sides of our triangles where the height is defined as the 45^o degree elevation will be more than 60^o, or in other words, the angle required to turn at each corner will be less than 120^o.  I do not know enough about spherical geometry to figure what that turn angle is on the plane tangent to the hemisphere where that turn occurs.  I just know it is not as violent as 120^o.

Keith

[Trostle]

Sum of the angles of a spherical triangle range from 180 to 540 degrees.  A point triangle would sum 180; a triangle expanded to become a great circle would have 540.  However. the pilot sees the triangle as if it were on a flat surface, so can enjoy the illusion that it is a 60 degree corner equilateral.

Hi Jim,

I really do not mean to contradict you, but----

Yes, the pilot only sees the "actual" (whatever that means) figures he is flying while the airplane on the on the surface of our hemisphere.  To the pilot, the model is always traveling on a plane tangent to the surface of the hemisphere. That plane of tangency is always changing as the model moves.  That can be seen only from the center of the hemisphere where the pilot is standing.  Now, when it comes to the triangle as described above by Tim where each of the legs of the triangle are 90^o great circle paths on the hemisphere, the pilot will actually see an equilateral triangle and see real 90^o corners, not the 60^o corners you suggest.  The turn at the outset is 90^o, just like the initial turn of a wingover, the turn at the top of the circle can be seen as an actual 90^o turn, and the pull out at the bottom is another 90^o turn, just like the exit of the wingover.  All of the sides are equal length, all of the turns are the same.

Keith

[Jim Thomerson]

My solid geometry book is hiding from me. When I find it I will remember more.   

[Ted Fancher]

Good stuff.  The Kernel and I have probably "discussed" this sort of thing once or twice over the several decades of our er...um...relationship.

I think it might be safe to say the following...which may not be measurable.  If the pilot can see all three corners of the equilateral triangle he is flying while using only peripheral vision (staring at the center of the figure) the figure he sees will include three 120 degree changes in the airplane's pitch axis generated path.  Each corner will be equidistant from the eyeballs, thus appearing as though drawn on the wall in front of him/her on a plane square to a point 22.5 above its (the unisex version of his/her) eyeballs.

Thus, the maximum size of a triangle that appears unilateral to the pilot will depend entirely on the width (in 360 degrees) of its peripheral vision and the length of the flying lines!

Or, maybe not.

Ted

p.s.  It's probably threads like this that make perfectly capable individual believe they couldn't possibly judge stunt!

[Tim Wescott]

p.s.  It's probably threads like this that make perfectly capable individual believe they couldn't possibly judge stunt!

As long as I can bring my laptop to work out the higher math, I'll be fine.

[Steve Helmick]

I originally clicked on the YouTube video because I thought it might be Howard doing some of his fine triangles. Then, I was wishing that ol' Pushpinder had drawn the equilateral triangles with radii instead of pointy points. Finally, I just thought the equilateral triangle diagram was so much better than most triangles I see down at the stunt circle, I just had to post the link. Sorry 'bout that.  Steve

[Brett Buck]

I originally clicked on the YouTube video because I thought it might be Howard doing some of his fine triangles. Then, I was wishing that ol' Pushpinder had drawn the equilateral triangles with radii instead of pointy points. Finally, I just thought the equilateral triangle diagram was so much better than most triangles I see down at the stunt circle, I just had to post the link. Sorry 'bout that.  Steve

   The spherical geometry is of academic interest, but what an equilateral triangle looks like is not really debatable, and certainly there can be no debate about how to judge or coach it. 3 tight corners and three straight equal straight lines and you get a 40. Now, if only anyone could reliably do it.

     Brett

[Jim Thomerson]

While judging, I have seen it done once, with an ARF Nobler.  Second flight not quite as good.

[RogerGreene]

Well, if all three sides do not equal 360d then you are either planted in the ground or flying through the triangle or some variance between! 

[Tim Wescott]

Now, if only anyone could reliably do it.

I recently pinned an equilateral triangle made of string to my office wall.  When I need to give my eyeballs a break from peering at the 'puter screen I swivel around in my chair and trace triangles with my finger.

I don't yet know how much it'll really help, but the two flights that I've been able to do since it went up have included better triangles.

[Steve Helmick]

When "dry flying" in the livingroom, I have trouble getting the second triangle as good as the first.  Steve

[Leo Mehl]

When "dry flying" in the livingroom, I have trouble getting the second triangle as good as the first.  Steve

I noticed that!

[phil c]

You gotta read the rule book.  The description says the top "corner" should be at 45deg.  The drawing shows the same.  Given the average 15-17ft. corner radius the actual point of the containing triangle is about that far outside the "corner", so the figure is actually about a 58deg"triangle.  The width will be about 50 degrees, smaller than the containing triangle, but larger than the 45deg elevation because of the corner radius.

This also shows that if someone makes a sharper corner, the height will stay the same, but the maneuver will be smaller.   A subtle change that might cause some judges to downgrade the maneuver because the width is noticeably less.  Conversely, a larger corner radius will make the maneuver smaller in width, closer to the specified 45 degrees and confusing things.

[Brett Buck]

You gotta read the rule book.  The description says the top "corner" should be at 45deg.  The drawing shows the same.  Given the average 15-17ft. corner radius the actual point of the containing triangle is about that far outside the "corner", so the figure is actually about a 58deg"triangle. 

  No, it doesn't. The top is supposed to be at the same 45 degrees as all the other maneuvers. The larger the radius, the shorter the sides get, but the maneuver size stays the same. This is why the top fliers maneuvers look better, and why Paul Walker's triangles consistent score well - because the straight sides are very well-defined as nice straight lines. Because the corner radius is small AND, there's very little "swoop" to get to the minimum radius. This is what matters.

    Making the maneuver oversize tends to also make the maneuver straight sections look longer and better-defined, as does making it physically larger (and the right angular dimensions) by using longer lines. This is why many people make the maneuvers larger than specified, because it improves the definition of the maneuver and makes the corners looks smaller by comparison. Not properly deducting for oversize maneuvers is a double-whammy as far as judging goes.

     Note that the corner radius issue results in the hourglass being much narrower than the triangle. Crossing the intersection at 45 degrees has zero radius, a typical maneuver corner is around 13 feet, so the hourglass is defined to be something like 21-22 feet narrower than the triangle. If you try to make it the same width, the angle become much more acute.

    Brett

[Mike Keville]

Speaking as a (former) Nats-level judge, I was often dismayed to see the 2nd leg of the Triangles (and 3rd leg of the Hourglass) flown in a nearly vertical dive.  Of course those were downgraded accordingly.

Without mentioning names here, I've seen four guys who consistently fly TRUE Triangles and Hourglasses rather than near-vertical dives at the conclusion.

(Don't bother asking.)

[Brett Buck]

Speaking as a (former) Nats-level judge, I was often dismayed to see the 2nd leg of the Triangles (and 3rd leg of the Hourglass) flown in a nearly vertical dive.  Of course those were downgraded accordingly.

    My observation has been somewhat different. What I see on the hourglass, time after time, is that the first leg is about right or too shallow, but FAR too short. The second corner is usually far too little, resulting in a z-shaped climb. The third corner is then nearly directly overhead, the pilot tries to tun it back to the right, but can't, stalls out or runs out of line tension, resulting in either an shaped dive, or, as you note, nearly straight down descent.

    Brett

[Mike Keville]

Saw that.  We are in agreement - as usual.

[Brett Buck]

Saw that.  We are in agreement - as usual.

   The only reality is what shows up on the scoreboard...

    Brett

[Mike Keville]

   The only reality is what shows up on the scoreboard...

    Brett

==============================================
True.  Yet not all judges downgrade for those 'vertical dives' on the back half of the triangles and hourglass.

It is what it is.

(Your maneuvers, by the way, aren't in question here.)

[Trostle]

  Given the average 15-17ft. corner radius the actual point of the containing triangle is about that far outside the "corner", so the figure is actually about a 58deg"triangle. 

I am going to repeat what Brett said regarding the above statement, but say it in a different way.  The tops of the triangles are defined as the 45^o elevation and have nothing to do with the radius of any of the corners actually turned in the maneuver.

Keith

[Brett Buck]

I am going to repeat what Brett said regarding the above statement, but say it in a different way.  The tops of the triangles are defined as the 45^o elevation and have nothing to do with the radius of any of the corners actually turned in the maneuver.

  Phil is projecting the sides to come to a hypothetical "point" which may well be in the 58 degrees range. Apparently the word "corner" is somehow confusing as is the drawing with 45 degrees, a dotted line, and it going to the top of the maneuver, along with every single thing ever said about triangle in the extensive discussions.

   Brett

     

[Tim Wescott]

   My observation has been somewhat different. What I see on the hourglass, time after time, is that the first leg is about right or too shallow, but FAR too short. The second corner is usually far too little, resulting in a z-shaped climb. The third corner is then nearly directly overhead, the pilot tries to tun it back to the right, but can't, stalls out or runs out of line tension, resulting in either an shaped dive, or, as you note, nearly straight down descent.

I'm currently struggling with an hourglass that is nicely symmetric around a vertical great circle, intersects at about 45 degrees, and doesn't make it all the way up to vertical -- so I have a smaller triangle on top than on the bottom.

[Tim Wescott]

I recently pinned an equilateral triangle made of string to my office wall.  When I need to give my eyeballs a break from peering at the 'puter screen I swivel around in my chair and trace triangles with my finger.

And it's been solid rain (or snow) from when I posted that until yesterday.  And I only put in three flights yesterday, but on two out of three flights my triangles were great -- and on the one flight where they weren't, the first one was shallow but the second one was just dandy.

I guess this means I need to paper my walls with squares and circles, now, too.