Design Calculation of a Cone

Hi Buddy,

Pls find the calac formulas below.

Lets make it this way: D=500 d=25 h=1000 s=Sqrt (R-r)²+h² Volume= 0.2618h(D²+Dd+d²) Area=1.5708 s (D+d) In your case: Volume=68,886 Area=847,609

Regards,

AS

Now wrap this cone with "22" AWG wire, spiraling jam-packed from base to the top. What will be the rate of change in radius of this spiral around the" cone" and also rate of change in height from "base" to "top" as this wire spirals to the top! Differential equations will be very helpful.

Gidday gemrafi,

First off I havn't conversed with you before so welcome to CR4 and I hope you get as much out of it and enjoy it as much as I.

What we have here is a classic example of a two cone problem as shown in the diagram.

The first thing we need to do is calculate the height of the small cone at the top of the large cone. Since the two cones are of identical profile the ratio of their height to ratio is identical. Given that

• R_S = Small Radius
• R_L = Large Radius
• H_L = Height of large cone
• H_S = height of small cone
• H_O = height overall

Now that we have the height of the two cones it is a simple matter of calculating the volume and surface area of both cones and subtracting the volume and surface area of the smaller cone.

The surface area is more complex and needs to be broken up into several sections as follows

• A_C = Surface of large cone face
• A_S = Surface area small cone face
• B_L = Surface are large base
• B_S = Surface area small base

To calculate these we need to know the length of the cord down the side of each cones

• C_L = Cord of large cone
• C_S = Cord of small cone

We can use Pythagoras's theorem to calculate the two cords as follows

The face of the cone is actually a segment of a circle that has the radius of the cord of the circle. The portion of this circle is defined by the ratio circumference of the base of the cone divided by the circumference of the circle subscribed by the cord of the cone. This ratio we will call C_R and it can be calculated as follows

We can now calculate the area of the face of the large and small cones as well as the areas of the bases of the large and small cones using the ratio C_L and the usual formula for the area of a circle

Area of large cone face

Area of small cone face

Area of large base

Area of small base

We can now calculate the total surface area of the truncated cone as

So the answer to your question is

• Volume = 68.88 Liters
• Surface Area = 1.044 square metres

The answer is fairly long winded but I have included as much of the workings of the calculations so that you can understand the derivation rather than just quoting the formula.

I and not sure how Avadiar siva got a volume that was out by a factor of 10 but the discrepancy in the surface area was caused by omitting the areas of the two bases that form the end of truncated cone.

Good effort masu, thanks with my rate "Good Answer".

Send me your email address and I will send you a working solution in Excel.

mark_howell@floridacrystals.com

sir

thanks for your idea about cone. what my question is suppose we have the 5mm thickness of plate of 2mx 2m dimension.that time we cut the plate for cone design.

how i am marked for cutting and what dimension for cutting.what is the the calculation

thanking you

rafic

thanks for your idea about cone. what my question is suppose we have the 5mm thickness of plate of 2mx 2m dimension.that time we cut the plate for cone design.

how i am marked for cutting and what dimension for cutting.what is the the calculation.

If I understand your follow-up question correctly, you are asking for the flat pattern layout needed to form the curved surface (only) of the hollow cone. Is that right? If so, take the two "cords" (sometimes known as "elements" of the cone) that MASU has calculated, and use them as radii on your flat sheet. Begin your swing of the large one at one corner of your original plate, with the center point on one of the edges leading to it; this provides one of your join lines and usually maximizes the useful area of your sheet. Swing the second radius from the same center. Next, you need the angle between this edge and the other join; it will be 83.181 degrees if I've done my math correctly. MASU's C-sub-l is 1,081.91 mm (NOT 1,81.91 as shown, I believe). Divide this into the radius of the large base circle (250 mm) to get a ratio of 0.231: this is the fraction of a circle that you need. Multiply 0.231 x 360 degrees to get 83.186 degrees. You may add a width for overlapping or other use as needed. Try this with paper, at smaller scale, to confirm the method! When you roll up this [almost] quarter circle, you will have your cone.

Here is what the developed shape looks like

For each of the measurements you need to use the values from my first post which gives

• RS = R_S = 12.5 mm
• RL = R_L = 250 mm
• CS = C_S = 54.9 mm
• CL = C_L = 1,01,91 mm
• Angle = C_R x 360° = 83.19°

By the way I just noticed that in my original pose #2 the value I have given for CL is

C_L = 1,81.91

This calculation and value should actually be:

Sorry about that, I noticed it at the time but forgot to go back and correct it before posting it

If you have any further queries please don't hesitate to ask.