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3 years ago

Why should a toolpath be verified on the screen of a CAM system prior to creating the program code?

Engineering

1 answer:

NISA [10]3 years ago

5 0

Answer:

The tool's trajectory in a CAM program refers to the places where the tool will be during the work. It is important to review it before generating the program for the following reasons

1. analyze the machining strategy and identify which one is better for each piece.

2.Avoid the collision of the tool holder with the work piece.

3.Avoid the shock of the tool with the piece.

4. Prevent the collision of the tool with elements that are not displayed on the CAM such as clamping flanges or screws.

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Ben leads a team of a few engineers at a robotics firm. A couple of them would like to improve their skills by taking additional

Anit [1.1K]

Answer:

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Explanation:

4 0

2 years ago

6.28 A six-lane freeway (three lanes in each direction) in rolling terrain has 10-ft lanes and obstructions 4 ft from the right

dimulka [17.4K]

Answer:

Assume Base free flow speed (BFFS) = 70 mph

Lane width = 10 ft

Reduction in speed corresponding to lane width, fLW = 6.6 mph

Lateral Clearance = 4 ft

Reduction in speed corresponding to lateral clearance, fLC = 0.8 mph

Interchanges/Ramps = 9/ 6 miles = 1.5 /mile

Reduction in speed corresponding to Interchanges/ramps, fID = 5 mph

No. of lanes = 3

Reduction in speed corresponding to number of lanes, fN = 3 mph

Free Flow Speed (FFS) = BFFS – fLW – fLC – fN – fID = 70 – 6.6 – 0.8 – 3 – 5 = 54.6 mph

Peak Flow, V = 2000 veh/hr

Peak 15-min flow = 600 veh

Peak-hour factor = 2000/ (4*600) = 0.83

Trucks and Buses = 12 %

RVs = 6 %

Rolling Terrain

fHV = 1/ (1 + 0.12 (2.5-1) + 0.06 (2.0-1)) = 1/1.24 = 0.806

fP = 1.0

Peak Flow Rate, Vp = V / (PHV*n*fHV*fP) = 2000/ (0.83*3*0.806*1.0) = 996.54 ~ 997 veh/hr/ln

Vp < (3400 – 30 FFS)

S = FFS

S = 54.6 mph

Density = Vp/S = (997) / (54.6) = 18.26 veh/mi/ln

7 0

3 years ago

End A of the uniform 5-kg bar is pinned freely to the collar, which has an acceleration = 4 m/s2 along the fixed horizontal shaf

sergiy2304 [10]

Answer:

Fa = 57.32 N

Explanation:

given data

mass = 5 kg

acceleration = 4 m/s²

angular velocity ω = 2 rad/s

solution

first we take here moment about point A that is

∑Ma = Iα + ∑Mad ...............1

put here value and we get

so here I = ( $\frac{1}{12}$ ) × m × L² ................2

I = ( $\frac{1}{12}$ ) × 5 × 0.8²

I = 0.267 kg-m²

and

a is = r × α

a = 0.4 α

so now put here value in equation is 1

0 = 0.267 α + m r α (0.4) - m A (0.4)

0 = 0.267 α + 5 (0.4α) (0.4 ) - 5 (4) 0.4

so angular acceleration α = 7.5 rad/s²

so here force acting on x axis will be

∑ F(x) = m a(x) ..............3

a(x) = m a - m rα

put here value

a(x) = 5 × 4 - 5 × 0.4 × 7.5

a(x) = 5 N

and

force acting on y axis will be

∑ F(y) = m a(y) .............. 4

a(y) - mg = mrω²

a(y) - 5 × 9.81 = 5 × 0.4 × 2²

a(y) = 57.1 N

total force at A will be

Fa = $\sqrt{a(x)^2+a(y)^2}$ ...............5

Fa = $\sqrt{(57.1)^2+(5)^2}$

Fa = 57.32 N

3 0

3 years ago

What is the primer coating that protects the metal from rusting on a Aftermarket part.. Flat Primer

Anarel [89]

Primer sealer is the correct option choice for your question! Hope this helps

7 0

3 years ago

Assume that 50,000 yd3 of the soil from a local borrow pit in Rhode Island, with a water content of 13% and dry unit weight of 8

Inga [223]

Answer:3000 yd3

Explanation:

so i got the 50000 divided it by 100 times that by 13 and got 4500 yd3

so now the soil is at 9 so i need to get the 4 percent back up so i yet again got 50000 yd3 and divided by 100 and times it by 9 to get 1500 yd3 to then minus my 1500 and 4500 yd3 and got 3000 yd3 i hope it is right.

4 0

3 years ago