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Burka [1]
3 years ago
9

An equal-tangent vertical curve is to be constructed between grades of -2% (initial) and 1% (final). The PVI is at station 110 0

0 and at elevation 420 ft. Due to a street crossing the roadway, the elevation of the roadway at station 112 00 must be at 424.5 ft. Design the curve, and determine the elevations and stations for PVC and PVI.

Engineering
1 answer:
Mila [183]3 years ago
4 0

Answer:

The curve length (<em>L</em>) will be = 1218 ft

The elevations and stations for PVC and PVI

a. station of PVC = 103 + 91.00

b. station of PVI = 116 + 09.00

c. elevation of PVC = 432.18ft

d. elevation of PVI = 426.09ft

Explanation:

First calculate for the length (<em>L</em>)

To calculate the length, use the formula of "elevation at any point".

where, elevation at any point = 424.5.

and ∴ PVC Elevation = (420 + 0.01L)

Then, calculate for Station of PVC and PVI and elevation of PVC and PVI

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3 0
3 years ago
A 3-in-thick slab is 10 in wide and 15 ft long. The thickness of the slab is reduced by 20% and width increases by 3% in a hot-r
xxMikexx [17]

Answer: l = 2142.8575 ft

v = 193.99 ft/min.

Explanation:

Given data:

Thickness of the slab = 3in

Length of the slab = 15ft

Width of the slab = 10in

Speed of the slab = 40ft/min

Solution:

a. After three phase

three phase = (0.2)(0.2)(0.2)(3.0)

= 0.024in.

wf = (1.03)(1.03)(1.03)(10.0)

= 10.927 in.

Using constant volume formula

= (3.0)(10.0)(15 x 15) = (0.024)(10.927)Lf

Lf = (3.0)(10.0)(15 x 15)/(0.024)(10.927)

= 6750 /0.2625

= 25714.28in = 2142.8575 ft

b.

vf = (0.2 x 0.2 x 3.0)(1.03 x 1.03 x 10.0)(40)/(0.024)(10.927)

= (0.12)(424.36)/0.2625

= 50.9232/0.2625

= 193.99 ft/min.

4 0
3 years ago
Give me the description of - Feedforward control loops with an example.
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Answer:

Feedforward basically configured and used mainly to avoid errors in a control system entering or disrupting a control loop

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Feedforward basically configured and used mainly to avoid errors in a control system entering or disrupting a control loop. Although Feedforward control seems to be a very attractive idea, it imposes a high responsibility on both the system developer and the operator to examine and consider mathematically the effect of disruptions on the process concerned.

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4 years ago
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Read 2 more answers
Three communications channels in parallel have independent failure modes of 0.1 failure per hour. These components must share a
ozzi

Answer:

the MTTF of the transceiver is 50.17

Explanation:

Given the data in the question;

failure modes = 0.1 failure per hour

system reliability = 0.85

mission time =  5 hours

Now, we know that the reliability equation for this situation is;

R(t) = [ 1 - ( 1 - e^{-0.1t )³] e^{-t/MTTF

so we substitute

R(5) = [ 1 - ( 1 - e^{-0.1(5) )³] e^{-5/MTTF = 0.85

[ 1 - ( 1 - e^{-0.5 )³] e^{-5/MTTF = 0.85

[ 1 - ( 0.393469 )³] e^{-5/MTTF = 0.85

[ 1 - 0.06091 ] e^{-5/MTTF = 0.85

0.9391 e^{-5/MTTF = 0.85  

e^{-5/MTTF = 0.85 / 0.9391

e^{-5/MTTF = 0.90512

MTTF = 5 / -ln( 0.90512 )

MTTF = 50.17

Therefore, the MTTF of the transceiver is 50.17

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