Simple 2D drawing with manual calculation

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mrkhunj

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Hi everyone, i believe we can have formulae for this simple 2D drawing for everyone who don't learn engineering subject like me.
Given known value for fixed length (L) and diameter (Ø), need to calculate the max height (H). Diameter all same size.
1662778935812.png
thanks geniuses!!!
it will be great help for me(and others) since i dont know to use any drawing applications/softwares
 
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mrkhunj said:
Hi everyone, i believe we can have formulae for this simple 2D drawing for everyone who don't learn engineering subject like me.
Given known value for fixed length (L) and diameter (Ø), need to calculate the max height (H). Diameter all same size.
View attachment 33999
thanks geniuses!!!
it will be great help for me(and others) since i dont know to use any drawing applications/softwares
Click to expand...
What are the constraints?

Maximum height would be Φ1 + Φ2 + Φ3 with the constraint that every sphere must touch at least one other sphere.
 
Subhotosh Khan said:
What are the constraints?
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Good point! My initial visualization was that at least one circle must contact two others, with maximum height occurring when each circle contacts two others. We need more information. (H might be Φ, if L ≥ 3Φ.)

?
[imath]\;[/imath]
 
We might assume he meant it to look like his picture. So the bottom two circles could touch each other at one extreme and be far enough apart for the third circle to squeeze between them at the other extreme. You can't have [imath]L > 3\Phi[/imath] or [imath]L < 2\Phi[/imath]. I don't see it as an optimization problem, just the question of calculating [imath]H(L,\Phi)[/imath]. Under this interpretation [imath]H[/imath] would be max when [imath]L= 2\Phi[/imath] but calculating the formula for [imath]H(L,\Phi)[/imath] would be the interesting work. Also, to the OP, why not use [imath]d[/imath] or [imath]D[/imath] for the diameter instead of [imath]\Phi[/imath]?
 
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@mrkhunj here's a hint. I recommend that you mark the centre of each circle in your diagram. Join the 3 centres to make a triangle. With the addition of an extra line, in the right place, then you'd be able to use Pythagoras.
 
Subhotosh Khan said:
In "old" engineering drawings, diameter is often expressed as Φ .
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I'm familiar with the usage shown here,

The symbol or variable for diameter, ⌀, is sometimes used in technical drawings or specifications as a prefix or suffix for a number (e.g. "⌀ 55 mm"), indicating that it represents diameter. For example, photographic filter thread sizes are often denoted in this way.​

The symbol is distinct from phi, but quite close to the empty set symbol.

What I'm not sure of is whether that symbol is ever used as the OP uses it, as a variable name, particularly with a subscript. Wikipedia calls it a variable, but I can't find any examples of its use other than to label a number, as they show.
 
Good morning all, firstly this is not homework, but i seek google search for math genius help for any math questions. So here i am trying at the first math community following the search result.
Secondly, yes all spheres are touching. The two spheres at the bottom are actual objects laying flat on the ground surface. I am wondering if there are formulae to calculate the top height. I know we can use CAD / any drawing software or use measure tape.
Thanks guys for your expertise! I ma enjoying all your ideas, MATH is fun to learn
 
mrkhunj said:
am wondering if there are formulae to calculate the top height
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Yes there is a formula. You can obtain it by acting on the advice in post#7. Why not have a try? You can post an image of a pencil drawing if that's easiest for you.

You'll be aiming to finish up with two right angled triangles in the image by "adding an extra line". I assume that you know Pythagoras theorem?
 
Cubist said:
Yes there is a formula. You can obtain it by acting on the advice in post#7. Why not have a try? You can post an image of a pencil drawing if that's easiest for you.

You'll be aiming to finish up with two right angled triangles in the image by "adding an extra line". I assume that you know Pythagoras theorem?
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yes. this is what i got ;
H = Ø + (Ø2-[(L-Ø)/2]2)1/2
 
LCKurtz said:
We might assume he meant it to look like his picture.
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That would've have been one assumption (fixed L and H, as shown). The OP said the bottom circles are spheres "laying flat on the ground". I'm not sure what that means, either. ;)
[imath]\;[/imath]
 
To make my interpretation clear, in the original picture think of [imath]L[/imath] as variable and the box has no top. If [imath]L[/imath] increases the middle ball will fall until when [imath]L[/imath] is [imath]3D[/imath] when the balls are side by side and [imath]H=D[/imath]. However if L decreases the middle ball will rise until the two bottom balls touch each other when [imath]L=2D[/imath]. It is easy enough to calculate the height then as [imath]H = D(1+ \frac {\sqrt 3} 2)[/imath] from the geometry. This agrees with the result from the formula in post 11 and others of H = Ø + (Ø^2-[(L-Ø)/2]^2)^(1/2) if you take L = 2Ø. So it is just an example of an optimization problem where the max occurs at an end point.
 
LCKurtz said:
If L increases the middle ball will fall until when L is 3D when the balls are side by side and H=D. However if L decreases the middle ball will rise until the two bottom balls touch each other when L=2D.
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I completely agree. In that case L is variable whose value is bounded between 2D and 3D. However, the op declares:

mrkhunj said:
Given known value for fixed length (L) and diameter (Ø),
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In that case H is also fixed from geometry. The above was not reinterpreted! The main FIND of the question - the maximum H does not make sense.
 
The OP also stated, "all spheres are touching". Without context, that statement was also ambiguous. (All spheres are touching what?) I think the OP has received what they wanted, but I hope the discussion has provided them with more than that.

:)
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