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2 years ago

What is the definition of a tolerance on a dimension typically found on technical drawings?

1 answer:

7 0

Tolerance is the acceptable amount of dimensional variation that still allows a part to perform as designed.

Any process will have variation and depending on the severity of the function some tolerance will be very small. For example the sheet metal thickness on portion of a space shuttle will have a much tighter tolerance than the thickness of a piece of lumber to build a house. Tighter tolerance of processes typically are related to more process control (e.g. money) thus designs should be fully vetted with process team before placing on a drawing.

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Give me source code of Simple openGL project. ( without 3D or Animation) simple just.

Ivan

Answer:

Use GitHub or stackoverflow for this answer

Explanation:

It helps with programming a lot

4 0

2 years ago

cThe Mars Rover Spirit got stuck in the Martian sand. The wheels kept slipping. Attempts to free it were futile. Discuss the typ

IgorC [24]

Answer:

Improved/ advanced types of Actuators include servo systems, create a large range of actuator motion in response to the changing needs of the operational environment or process.

Actuators are local or automated suppliers of working motion.

Hydraulic and air cylinders can be classified as single-acting cylinders, meaning that the energy source result in movement in one direction and a spring is used for the other direction.

Explanation:

An actuator control system is referred to as any electronic, electrical, or electromechanical system often used to activate an actuator, control the direction as well as extent and duration of its output. Actuator control systems could take the form of extremely simple, manually-operated, start-and-stop stations, either sophisticated or programmable computer systems. The more improved/ advanced types include servo systems that produce a large range of actuator motion in response to the changing needs of the operational environment or process. This type of actuator control system uses an interface arrangement that assimilates feedback from the process or mechanism and adjusts the actuator in the right way. Most actuator systems will include at least a set of travel limits that prevent the actuator destroying itself or the secondary mechanism.

Actuators are local or automated suppliers of working motion. They are used to changes, adjust, or move a secondary mechanism, where a physical operator cannot intervene directly. They are denoted by a large range of varying types using electrical and electromagnetic, hydraulic, or pneumatic power sources to create linear or rotary outputs. One element they all have in common is the actuator control system used to start, stop, and adjust the range, speed, and duration of the working motion.

Actuators can produce a linear motion, rotary motion or oscillatory motion which means they can create motion in one direction, in a circular motion or in opposite directions at regular intervals. Hydraulic and air cylinders can be classified as single-acting cylinders, meaning that the energy source result in movement in one direction and a spring is used for the other direction.

7 0

3 years ago

• Build upon the results of problem 3-85 to determine the minimum factor of safety for fatigue based on infinite life, using the

Rudik [331]

Answer:

minimum factor of safety for fatigue is = 1.5432

Explanation:

given data

AISI 1018 steel cold drawn as table

ultimate strength Sut = 63.800 kpsi

yield strength Syt = 53.700 kpsi

modulus of elasticity E = 29.700 kpsi

we get here

$\sigma a$ = $\sqrt{(\sigma a \times kb)^2+3\times (za\times kt)^2}$ ...........1

here kb and kt = 1 combined bending and torsion fatigue factor

put here value and we get

$\sigma a$ = $\sqrt{(12 \times 1)^2+3\times (0\times 1)^2}$

$\sigma a$ = 12 kpsi

and

$\sigma m$ = $\sqrt{(\sigma m \times kb)^2+3\times (zm\times kt)^2}$ ...........2

put here value and we get

$\sigma m$ = $\sqrt{(-0.9 \times 1)^2+3\times (10\times 1)^2}$

$\sigma m$ = 17.34 kpsi

now we apply here goodman line equation here that is

$\frac{\sigma m}{Sut} + \frac{\sigma a}{Se} = \frac{1}{FOS}$ ...................3

here Se = 0.5 × Sut

Se = 0.5 × 63.800 = 31.9 kspi

put value in equation 3 we get

$\frac{17.34}{63.800} + \frac{12}{31.9} = \frac{1}{FOS}$

solve it we get

FOS = 1.5432

6 0

2 years ago

Why can you anodise Aluminium and Magnesium alloys?

Anastasy [175]

Explanation:

Anodizing :

Anodizing is the surface protection process from the environment.As we know that due to external environment surfaces get corrodes .By using anodizing process the outer surface of material coated by using different type of coating material.

As the name stand that in the anodizing process there will be anode and oxygen.in this process oxidation of material take place .

Oxides of aluminium and magnesium are stable that is why they anodized by this process.

4 0

3 years ago

A conical enlargement in a vertical pipeline is 5 ft long and enlarges the pipe diameter from 12 in. to 24 in. diameter. Calcula

makkiz [27]

Answer:

F_y = 151319.01N = 15.132 KN

Explanation:

From the linear momentum equation theory, since flow is steady, the y components would be;

-V1•ρ1•V1•A1 + V2•ρ2•V2•A2 = P1•A1 - P2•A2 - F_y

We are given;

Length; L = 5ft = 1.52.

Initial diameter;d1 = 12in = 0.3m

Exit diameter; d2 = 24 in = 0.6m

Volume flow rate of water; Q2 = 10 ft³/s = 0.28 m³/s

Initial pressure;p1 = 30 psi = 206843 pa

Thus,

initial Area;A1 = π•d1²/4 = π•0.3²/4 = 0.07 m²

Exit area;A2 = π•d2²/4 = π•0.6²/4 = 0.28m²

Now, we know that volume flow rate of water is given by; Q = A•V

Thus,

At exit, Q2 = A2•V2

So, 0.28 = 0.28•V2

So,V2 = 1 m/s

When flow is incompressible, we often say that ;

Initial mass flow rate = exit mass flow rate.

Thus,

ρ1 = ρ2 = 1000 kg/m³

Density of water is 1000 kg/m³

And A1•V1 = A2•V2

So, V1 = A2•V2/A1

So, V1 = 0.28 x 1/0.07

V1 = 4 m/s

So, from initial equation of y components;

-V1•ρ1•V1•A1 + V2•ρ2•V2•A2 = P1•A1 - P2•A2 - F_y

Where F_y is vertical force of enlargement pressure and P2 = 0

Thus, making F_y the subject;

F_y = P1•A1 + V1•ρ1•V1•A1 - V2•ρ2•V2•A2

Plugging in the relevant values to get;

F_y = (206843 x 0.07) + (1² x 1000 x 0.07) - (4² x 1000 x 0.28)

F_y = 151319.01N = 15.132 KN

6 0

2 years ago