Project Jam

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hansellitis

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Oil Company Z has decided to build a new pumping station. This station is in the middle of a swamp and they need a solution to construct a roadway across the swamp. Their construction crew has recommended that a bridge in the shape of a simple arc will be sufficient to complete the roadway.
The bridge must span a linear distance of 500 feet from points we will call C and D. The difference in elevation between C and D is 20 feet with D being the higher point. The bridge must connect smoothly with the existing roadways at points C and D. At the C side of the swamp the roadway has a 7% grade (slope of the roadway is 0.07) and at the D side the grade is 4% (slope of –0.04). C is at swamp level. A diagram showing a representation of what the bridge shape will look like is below in figure 1.


One engineer proposed to use a parabolic arc that connected smoothly at D and had the equation y=0.12x-(1.6/10^4)x^2, however this arc has a slope at C of 0.12 which is not smooth since the slope of the roadway at C is 0.07.


Use a cubic equation to model the bridge span and find a formula which is smooth at both C and D.

http://spot.pcc.edu/~jkissick/spring/ma ... oject1.pdf
There's a link to the entire problem, but all I need help with is using a cubic equation to model the bridge span and find a formula which is smooth at both C and D.

I don't know if you guys help with things like this, because I am a n00b here. But I am stuck and am in desperate need for help!


Here's the huge problem, one of our group leaders was got far ahead of us on the project and came up with the correct formula: -200(10^-9)(x^3)+40(10^-6)(x^2)+.07x

Then he dropped out of the class a few days later, leaving us with 0 work shown. And I can't figure out how he got that formula (But I know its right because it works out for everything).

NOTE: The teacher allows us to talk with other students/teachers about projects, however we are not allowed to talk with the teacher.
 
WORK:
In order for the bridge to smoothly transition to the road bed on each end of the bridge the end of the bridge at point C will need to match the 7% grade of the road and the point at the D side of the bridge will have to meet the 4% down hill grade. In order to meet both of these requirements the following cubic equation was created.


-200(10^-9)(x^3)+40(10^-6)(x^2)+.07x (again one of our teammates who dropped out gave us this problem so we don't have the work)

In order to identify where to dredge the channel under the bridge, we input two values for y into the calculator, graph these points and find the intersection points. The following are the y values:

f(x)=-200(10^-9)(x^3)+40(10^-6)(x^2)+.07x
f(x)=21.5

With these values the intersection points are found to be at 388.04245 ft and 440.64834 ft. These values are the distance from Point C in figure 2. The channel can be dredged between these points the barge holding the crane will fit under the bridge anywhere between these points.

With this information the construction crew can construct a bridge that meets all of the necessary requirements to provide a solution that the barge can easily pass under.


We did everything else perfect. It's just we don't know how our teammates found that formula...
 
hansellitis said:
We did everything else perfect. It's just we don't know how our teammates found that formula...
Click to expand...
What's the "everything else" that you did...? And why would you think that we would know the steps and reasoning behind whichever other part that your teammate did...?

Also, what objection do you have to the worked solution provided in the other thread?

Please be specific. Thank you! :D

Eliz.
 
stapel said:
hansellitis said:
We did everything else perfect. It's just we don't know how our teammates found that formula...
Click to expand...
What's the "everything else" that you did...? And why would you think that we would know the steps and reasoning behind whichever other part that your teammate did...?

Also, what objection do you have to the worked solution provided in the other thread?

Please be specific. Thank you! :D

Eliz.
Click to expand...

I don't need you guys to understand the project. I just need help finding the cubic equation. And someone already did that in the other forums.

All the other work consisted of typing up a 3 page paper with all the relevant information that our instructor wanted. He said everything was perfect, except he wanted work us to show work on the formula f(x)=-200(10^-9)(x^3)+40(10^-6)(x^2)+.07x. And like I've already stated, we've tried really hard to figure out how our ex-teammate found this. But we have been unable to figure it out. So that's why I'm requesting for help on this matter.
 
That equation your teammate derived is derived the same as the one in the other post. For goodness sake, check it out. The derivation is given. I don't know what else we can do. The only thing different is that 1/25 instead of -1/25 was used for the outgoing slope.

Set up your equations and find a and b. c and d are easy.
 
"That equation your teammate derived is derived the same as the one in the other post."

Well he used integrating... And we have not yet learned that in our class...


So this is what I got so far:

f(x)=ax³ + bx² + cx+d
f(0)=a0³ + b0² + c0+d -->f(0)=d, so d=0

f'(x)=3ax² + 2bx + c
f'(0)=3a0² + 2b0 + c -->f'(0)=c, so c=0.07

f(x)=ax³ + bx² + (0.07)x
f(500)=a(500)³ + b(500)² + (0.07)(500) -->125,000,000a+250,000b+35=20 -->125,000,000a+250,000b=-15

f'(x)=3ax² + 2bx + 0.07
f'(500)=3a(500)² + 2b(500) + 0.07 -->750,000a+1,000b+0.07=-0.04 -->750,000a+1,000b=-.11



a=(-0.11-1,000b)/750,000

(-0.11-1,000b/750,000)x³ + bx² + 0.07x=20

(-.11(10^-4)x³/7.5)-bx³/750 +b x²+0.07x-20=0

(-.11(10^-4)x³/7.5)-b x³/750 +b x²+0.07x-20=b(x³/750+x²)
((-.11(10^-4)x³/7.5)-0.07x-20)/ (x³/750+x²)=b
((-.11(10^-4)(500)³/7.5)-0.07(500)-20)/ ((500)³/750+(500)²)=b
b=-1.3200000352E-4

a=(-0.11-1,000(-1.3200000352E-4))/750,000 --> 2.93333380267E-8

So
(2.93333380267E-8)(x³)+(-1.3200000352E-4)(x²)+0.07x



I obviously did something wrong. But not sure where, anyone know what went wrong?

Thanks
 
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