Cones Expanse






When you want to cut incline pipe you need a drawing that using perspective drawing, but if you want to create cones from a sheet you don’t need a perspective drawing because this build more simple than incline cutting pipe.

If you want to build a cones like the above drawing you just need what is the top diameter and what the length of your cones, you just create a part of circle with the top diameter. The complete cones will need 1/3 of round circle.

1/3 of round circle is 1200 so you can cut the sheet to create angle 1200 then you create a circle with the diameter of the cones you need on the angle. Then you create again the part circle on the bottom part by using the same center with the top circle.

For example you have a form like the above sample then you can cut the bottom part because bottom should be a hole.

Comments

Post a Comment

Popular posts from this blog

Mathematical Language

Image
--> In mathematical language, the propagation problem is known as an initial-value problem (Fig. Mat-2). Schematically, the problem is characterized by a differential equation plus an open region in which the equation holds. The solution of the differential equation must satisfy the initial conditions plus any “side” boundary conditions. Fig. Mat-1 The description of phenomena in a “continuous” medium such as a gas or a fluid often leads to partial differential equations. In particular, phenomena of “wave” propagation are described by a class of partial differential equations called “hyperbolic,” and these are essentially different in their properties from other classes such as those that describe equilibrium (“elliptic”) or diffusion and heat transfer (“parabolic”). Fig. Mat-2 Prototypes are: 1. Elliptic. Laplace ’s equation d 2 u/dx 2 + d 2 u/dy 2 = 0 Poisson’s equation d 2 u/dx 2 + d 2 u/dy 2 =g(x,y) These do not co...
Read more

Statistic Problem

1. Known the number of ball n(s) = 100 Ask: The change take a ball with number divisible by 5 but not divisible by 3 if taken randomly. Answer: For example A = ball with number divisible by 5 but not divisible by 3 n(A) = 14, namely 5, 10, 20, 25, 35, 40, 50, 55, 65, 70, 80, 85, 95 and 100 P(A) = n(A)/n(s) = 14/100 = 7/50 2. Known: One badminton team consist of 5 members, will decide 2 people play a single players, and 2 pairs to play double. Ask: How many option can be made. First Way Solution: For example 5 people of badminton team are A, B, C, D, and E. Then we choose A and B to play single. Option Single Double 1 A, B CD, AE 2 A, B CE, AD 3 A, B  DE, AC 4 A, B CD, BE 5 A, B CE, BD 6 A, B DE, BC The option number when A and B choose as single player are 6 option. The way to choose player as single player are 10, namely AB, AC, AD, AE, BC, BD, BE, CD, CE, and DE. From one option of single player will make 6 option so for all way to choose si...
Read more

Mathematic Principle

The basic problems of the sciences and engineering fall broadly into three categories: Steady state problems. In such problems the configuration of the system is to be determined. This solution does not change with time but continues indefinitely in the same pattern, hence the name “steady state.” Typical chemical engineering examples include steady temperature distributions in heat conduction, equilibrium in chemical reactions, and steady diffusion problems. Eigenvalue problems. These are extensions of equilibrium problems in which critical values of certain parameters are to be determined in addition to the corresponding steady-state configurations. The determination of eigenvalues may also arise in propagation problems. Typical chemical engineering problems include those in heat transfer and resonance in which certain boundary conditions are prescribed. Propagation problems. These problems are concerned with predicting the subsequent behavior of a system from a knowledge of the in...
Read more