Random part of the day: Panel Curved 3X13X2

I didn’t realise so many sets had this part.

Anyone know how many it takes to make a circle? It would be gigantic!

Cool! I just got this piece in the McLaren Senna GTR last week.

Didn't realise the only set with these in Dk St Gray was the Osprey.

Wondering just how big a tub of these is sitting in the factory...

I wonder how this would look as a hammock

@MeisterDad:
You’re probably looking at a few dozen parts making a circle that’s over 3’ in diameter. However, because both ends have the same style connection, you’d also need twice as many 2L beams and 3L pins to link them.

@PurpleDave said:
" @MeisterDad :
You’re probably looking at a few dozen parts making a circle that’s over 3’ in diameter. However, because both ends have the same style connection, you’d also need twice as many 2L beams and 3L pins to link them."
I don’t know much about Technic pins, but isn’t there one that would allow you to create a chain of this RPotD without interstitial beams? I realise that that could mean that the curved panels would have to be staggered (zig-zagged) with one layer above and one below. If you could create a chain of these panels that way, with precise measurements of the panel, you can calculate the size of circle a chain of them would form.

You would need the distance between the centre of each connector, i.e. the chord length or ‘C’, and the maximum height between the chord and the panel or ‘H’ (the height at C/2 from the centre of the connector). The formula is:
Radius = H/2 + (C^2)/8H

The diameter is obviously twice the radius.

With the radius and C, you can work out how many panels are needed to form the circle.

Simples!

I'm mildly interested to note that 80013 is the ONLY non-technic set to include this piece, to date.

@Zander:
Yes, there are 2L pins. But if you have the ability to make such precise measurements and calculate the size of a circle formed by these (assuming the curve follows a consistent radius from pin to pin*), you should be able to factor in how the use of 2L beams affects the shape and size.

*There are two possible ways in which this does not follow a consistent radius. The shell itself may have a variable curve, and the tabs with the pin holes may not follow the curve at all. Either way, the pin holes are offset from the curved surface, which is going to complicate the math involved. This is a case where it would be a lot easier to solve it by practical means. Get enough to make two circles, and if you used beam extensions, you can lay one on top of the other and rotate it half a shell in one direction. Then you can use the shell from one circle as a guide to true up the joint on the other circle. Alternate your way around, and you’ll have everything close enough for government work in short order. It’s also worth noting that these may not form a true circle. It’s been noted on several occasions that parts that look like they should often end up off by a significant fraction of one part. For this part, you can fudge the shape of the circle quite a lot by either expanding or contracting the angle of each joint just a tiny bit.

After further consideration I must postulate that, by design, there is a stud length which can join the two pinholes beneath the arc of this part as a secant line, (chord) probably a 13L liftarm; otherwise why design the part with pin holes at both ends? This would suggest that a regular polygon with sides of length 13 can be constructed. The size of the circle in question would be the one circumscribed about this regular n-gon. This n would be the number of pieces needed to form a circle.

Further, using the part dimensions it seems that the greatest distance between the chord and the circumference of this circle is 2 studs (sagitta or segment height). I looked into this to see if there is a formula which can calculate the central angle of said circle knowing only these two lengths so that n can be determined and it turns out, no, there isn’t. I haven’t used Newton-Raphson in almost 20 years but I should dust off my rough pad and do it sometime.

http://www.1728.org/newton.htm

@MeisterDad said:
"After further consideration I must postulate that, by design, there is a stud length which can join the two pinholes beneath the arc of this part as a secant line, (chord) probably a 13L liftarm; otherwise why design the part with pin holes at both ends? This would suggest that a regular polygon with sides of length 13 can be constructed. The size of the circle in question would be the one circumscribed about this regular n-gon. This n would be the number of pieces needed to form a circle.

Further, using the part dimensions it seems that the greatest distance between the chord and the circumference of this circle is 2 studs (sagitta or segment height). I looked into this to see if there is a formula which can calculate the central angle of said circle knowing only these two lengths so that n can be determined and it turns out, no, there isn’t."

Not a single formula, no. But it is calculable. Assuming the curvature of the part from pin hole to pin hole is constant (I don’t know that it is but let’s assume), the distance between the centre of the pin holes is the chord. The radius of the circle (which is also the distance from each vertex of the polygon to the centre of the polygon) can be determined by the method I described in my post above. You then have three lengths of the isosceles triangle with the side of the polygon (i.e. the chord) as its base and the radius of the circle as the length of each of the other two sides. From there , you can work out the angle between the radii using the Cosine Law. 360 divided by that angle will give you n, the number of sides of the polygon which is also the number of panels you would need.

@MeisterDad:
The problem isn’t whether or not there’s a regular polygon that can be constructed with 13L sides, but that _every_ regular polygon can be built with 13L sides, and this leaves you no closer to discovering the one that allows this curved panel to form a (near) perfect circle.

@ThatBionicleGuy said:
"I'm mildly interested to note that 80013 is the ONLY non-technic set to include this piece, to date."

Some would say that by including some Technic elements, that makes 80013 a Technic set by definition!

@Galaxy12_Import said:
" @ThatBionicleGuy said:
"I'm mildly interested to note that 80013 is the ONLY non-technic set to include this piece, to date."

Some would say that by including some Technic elements, that makes 80013 a Technic set by definition! "

That definition would make 80% of the modern sets Technic

@MeisterDad said:
"Anyone know how many it takes to make a circle? It would be gigantic!"
10 is looking pretty round. Diameter is 32 cm.

@chefkaspa said:
" @MeisterDad said:
"Anyone know how many it takes to make a circle? It would be gigantic!"
10 is looking pretty round. Diameter is 32 cm.
"

Hats off to you for an empirical assessment!

@PurpleDave True, it may be impossible to have a regular n-gon with 13L in there; there might be some fraction of a part that would be too much or too little. I presupposed that the designers used their CAD to determine that it was so, which may be false and may anyway be inconsequential to the purpose of the part.

@chefkaspa:
Just a foot? Seems way too small, but I don’t think I’ve ever owned one of these, and may be thinking too large when picturing it.