Need help with triangles

[NathanR]

Hi, I could use some help working out which triangles built from a mix of technic beams and bricks are actually valid (or at least, not going to damage or strain the parts when they're assembled).  

For example, if you stack a load of 1x16 technic bricks, then attach a beam at an angle of about 22.6 degrees, every 14th pin hole on the beam lines up with a pin hole in a brick. Are there any other "magic angles" where the rotated beam's pin holes line back up with the grid? And if so, how can you calculate what they are, since 6 vertical bricks = a beam of 7 pin holes, and this seems to throw Pythagoras' law?

I'm also interested in calculating more "unusual" triangles - for instance, what if the stack of bricks includes one or two plates, or what if the bricks change to 1x1 technic bricks (i.e. pin holes with a half stud offset).  Is there a list of valid triangles, or a way to calculate what will work?  And what are the tolerances before such a triangle starts popping the bricks apart?

My specific use case is in a scale model of the launch pad for the Apollo moon rocket, one of the lower levels has to be 8 bricks and 2 plates high (plus or preferably minus 1-2 plates), and I'm looking for an angled beam that will produce the sloping side.  Unfortunately, it can't be under too much strain, as a ~1m tall tower will sit on top and need a reasonably solid foundation.  I'm going crazy with trial and error in LDD and a few rough prototypes in real life - could some kind soul give me some advice?

 

[DugaldIC]

I'm not entirely sure if this will help but there's already a thread by @Didumos69 on something very similar.

 

[JonathanM]

Basic thing you need to know is a brick's height (excluding the stud) is 6/5ths of it's width. And there's 3 plates to a brick, so a plate is 2/5ths it's width high. Technic holes are the same distance apart as bricks are wide, thus to get pinholes to line up you stack 2 plates between 2 bricks. Then the holes are 2 'widths' on center (2 plates plus a brick is 2/5+2/5+6/5=2 'widths' higher).

So 8 bricks and 2 plates is 8*6/5+2/5 = 10 units high. So an 11L beam will be exactly vertical on that height. There's no Pythagorus triangle to match that, but there's probably an approximate one. What is the angle needed?

[Saberwing40k]

So, you need a lattice structure that tapers? That's going to be fun, as the base of the LUT actually tapers in 2 directions, as I have observed from seeing some models. My 2 cents would be to actually cheat, and use straight transparent supports, and have the trusses, or at least the tapered part, be non structural.

[Didumos69]

9 hours ago, DugaldIC said:

I'm not entirely sure if this will help but there's already a thread by @Didumos69 on something very similar.

 

In general you can produce 'gridded' triangles using Pythagorean triples. The smallest ones are (3,4,5) and (5,12,13). You can also make use of the kite-shapes defined by the incircles of these triangles. The incircles have radius 1 and 2 respectively.

The nice thing about the (5,12,13)-triangle is that the angle made by the red line segments (180 - 2 arctan(2/10) = 157.38 degrees) practically coincides with the angle of a #3 connector (157.5 degrees).

Edited March 5, 2018 by Didumos69

[NathanR]

@DugaldIC, @Didumos69, I'm not sure how applicable this is in my case, because you're working with beams only, but I've got to say wow... that geometry, those connections...  it's just mind-blowing. I'm going to have to spend a lot of time rereading that other thread and really thinking about everything you're doing! 

7 hours ago, Saberwing40k said:

So, you need a lattice structure that tapers? That's going to be fun, as the base of the LUT actually tapers in 2 directions, as I have observed from seeing some models. My 2 cents would be to actually cheat, and use straight transparent supports, and have the trusses, or at least the tapered part, be non structural.

I did consider building with Pythagorean quadruples at first, but they just gave me a headache (not just the maths, but figuring out the 3D rotations in LDD).  In the end I "cheated" by having it taper in one direction only, and stepped in the other.  On the long side, each new layer is set back one extra brick, while the short sides then become angled plains - the truss-structure there is just made up of 3,4,5 triangles.  The design is actually :"in-system", on the long edge the two lower layers slope inwards at 22.6 degrees, with every 14th pinhole lining up.  The other triangles on the long side are set a couple of plates lower, and need something like a 15.01L beam for the connection, hence my use of 2L axles and axle connectors to build the strut, so it can "stretch" a little.

If you're thinking it looks perfect, wondering what the problem is... it's too tall! I need to shave two bricks worth of height off that second layer. Otherwise, there's no way to get the tower's arms lining up with the right bits of the rocket.

9 hours ago, JonathanM said:

So 8 bricks and 2 plates is 8*6/5+2/5 = 10 units high. So an 11L beam will be exactly vertical on that height. There's no Pythagorus triangle to match that, but there's probably an approximate one. What is the angle needed?

Well, there are two triangles needed really, but I think only one has to be structural. The outer triangle (the corner of the tower) should be around 22.6 degrees, the other triangle (for the truss structure) should be something like 50 degrees. I'm not so fussy on the truss structure, just so long as the tower is well supported.

Edited March 5, 2018 by NathanR

[Shiva]

12 hours ago, NathanR said:

My specific use case is in a scale model of the launch pad for the Apollo moon rocket, one of the lower levels has to be 8 bricks and 2 plates high (plus or preferably minus 1-2 plates), and I'm looking for an angled beam that will produce the sloping side.  Unfortunately, it can't be under too much strain, as a ~1m tall tower will sit on top and need a reasonably solid foundation.  I'm going crazy with trial and error in LDD and a few rough prototypes in real life - could some kind soul give me some advice?

Your Launch Pad Elevator shaft is part of the foundation too, right?

Sadly I can not help with the triangles.

[Oliver 79]

11 hours ago, Didumos69 said:

In general you can produce 'gridded' triangles using Pythagorean triples. The smallest ones are (3,4,5) and (5,12,13)........

I just can't get my head around this. I tried making a 3, 4, 5 triangle in LDD and it doesn't fit together. What am I doing wrong? My maths is rubbish. So if someone can explain in simple terms, I'd be very grateful!

[Rudivdk]

Actually you should be counting the spaces "in between" the pinholes. Try making it with a 4l, 5l and 6l beam, it will fit perfectly.

4l beam has 3x "in between" pinhole, 5l beam has 4 and 6l beam has 5. Took me awhile to figure this out myself as well...

 

[Oliver 79]

Say what!.....

So basically its the Pythagorean triples plus one one each side?

 

[Didumos69]

16 minutes ago, Oliver 79 said:

Say what!.....

So basically its the Pythagorean triples plus one one each side?

Yes. You have to measure pin-hole-center to pin-hole-center.

[NathanR]

20 minutes ago, Oliver 79 said:

So basically its the Pythagorean triples plus one one each side?

Yep.  The corner of the triangle is the centre of one pin hole, the distance between two pin holes is one "unit". So 5 "units" = 6 pin holes making up the hypotenuse of the 3-4-5 triangle.

@Didumos69 Snap!

Edited March 5, 2018 by NathanR

[Oliver 79]

Oh it's so simple when you know how! Thanks guys. 

[NathanR]

Small question, when building triangles that "just about work", do you have any idea what the tolerance actually is?

I have a triangle that may work, 5M beam base, 11M beam hypotenuse, 8 brick+2plate vertical height, angle about 22.7 degrees.  Mecabricks quotes the pin hole separation as 8 units (LDU?), and on this scale the pin holes nearly line up, with a 0.257 unit offset.  Do you think this close enough to work without problems?

[Didumos69]

From my experience working with a tolerance of 0.05M in LDD never leads to problems in a real life build. I know 42070 uses a 5,11,12 right triangle in the roof, where the hypotenuse of a right triangle with base 5 and height 11 should actually be 12.04. Only when a structure is supporting some kind of drivetrain with meshing gears etc. I would advice zero tolerance.

Edited March 5, 2018 by Didumos69