Education Geometry - Expert help needed - Straight cone with plain

I need some help from a geometry expert here.
- this is not homework for college or something,
- I'll paypal 5 euro to the first person who can provide all the formules I need.

All my problems are concerned with a "conical cup" and the wraparound label that fits around it. The geometrical shape, I believe, is a straight cone with a plain (top off).

See this pdf file for the flat shape of the wraparound label for a conical cup. Note: this is just an example, I'm looking for formules to calculate these results.

Formules needed:
- How to calculate the three top left values, using only the three lower right values;
- How to calculate the three lower right values, using only the three top left values;

I also need the formules to calculate the three lower right values, when been giving the diameter of the upper and lower plain of the conical cup and the absolute height.

I already figured out that the height of the label (60mm in this example) is related to the absolute height of the cup and the difference in diameter (simple pythagoras), but am stuck with all the rest.

If anything is unclear, please don't hesitate to ask questions (this is hard to explain in text...).

 

That doesn't look too hard, but appearances are often deceptive with geometry! I'll have a bash some time tomorrow, see if I can work some of it out.

 

Thanks, I've already proven in an autocad program that bouth is possible. I can manually draw the shape using only the three values on the right bottom and the program itself can automatically produce this drawing when provided with only the upper/lower diameter and the absolute height. So each set of three values is in relationship to eachother without needing extra input.

I just can't extract the way it does it.

cheers

 

Are you aware that the shape is symmetrical given its a blueprint for a cone? or is that not made clear in the questions?

 

I'm not sure what you mean, yes the shape is symetrical, so is the cup/cone. That does not solve my problem though ;-).

 

Are you told that in the question though?
If the question is just the shape as seen in the sample (without both sides sets of figures) then you can't assume that its symmetrical.

If yes then draw a line down the middle and you can instantly prove its an equalaterial triangle. 60* angle in all 3 corners so of course the sides are all equal lengths solving 320 and 260. 60 then is a simply a simple subtraction.

Oops first version wrong

260^2-130^2=square root about 225
320- above answer give you 94.83

Then the whole lot can of course be done in reverse if starting from the top figures.

 

Again this is not a 'question' as in some test, this is just an example I drew up. It is definatly not always an angle of 60 degrees, so the 320 on top does not match the 320 on the side all the time. It is symetrical, but not equalaterial.

I'll try out the above formule tomorrow, but I think it will only work when its equalaterial, so maybe this is a poor example.

But thanks for your input.

 

This would be my way of getting the 320 measurement.



The 60 is obviously the large radius minus the small radius.

 

That's the problem with giving the simplified examples. You use shortcuts.

Providing the triangle is an isocles then you can always cut it in half along the line of symetry. You now have 2 right angle triangles. Then its a simple case of applying sohcahtoa and pythag as needed.

EDIT
zener made a fancy diagram much clearer

The third number number is found by getting the length of the centre red line (a)by pythag
a^2 - b^2 = a^2
then subtracting that from the radius provided

 

A = 2 x (SIN(D/2) x E)
B = E - F
C = E - (COS(D/2) x F)

Ran out of time to look at the others.

 

Since there is some confusion, I've made a better example diagram.

http://pdfcast.org/pdf/geometry-example-cup

I've been using my high-school geometry & excel and have been able to calculate the following values/dimensions:
RU, using R1 and A
RL, using R2 and A
H , using R1, R2 and A
W , using R1 and A
R1, using RU and A
R2, using RL and A
A , using RU and R1
A , using RL and R2

But what I most of all need, is to calculate W and Husing RU, RL and LH.

Any help there?

cheers, Michiel

 

I can do it but I need to know R1:

I have no idea how to type a Pi symbol so will use the word!

W = 2 x (SIN( ( ( ( ( 2 x RU x Pi ) / ( 2 x R1 x Pi ) ) x 360) / 2 ) /2) x R1)

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Edit: Just realised that ( ( ( ( 2 x RU x Pi ) / ( 2 x R1 x Pi ) ) x 360) / 2 ) is just finding A and you already had that formula.

 

nice gj sparky.. I think a lot like mucgoo xD

 

I am sure there must be a way to find the angle without knowing R1 or R2 but I have no idea how to find it!

 

pythagore and thales are your best friends here ; ) .... think of isosceles triangles. I'll try to write it down if I have 5 minutes

Just a question :

Which formulae do you want ? What measurement do you want to calculate ?

 

But what I most of all need, is to calculate W and H using RU, RL and LH.

 

RU and RL are diameter ?

 

Yes.

Think it might be about time to say why!

 

RU and RH can be anything ... I do not see how their value could affect W and H.

But it is easy to calculate W and H using A, R1 and R2.

A question ..... are you trying to bend something ?

 

Ru and rh are the radius of the shape once it wrapped around. So 2 x Pi x RH and 2 x Pi x RU will give you the length of the curved lines.

Think of the shape as a cone of paper with the tip cut off then opened out and laid flat.