[chris1080p]

Hey guys, all I need is to know how to derive an elliptical equation and a parabolic equation from the info given. Thanks in advance.


A new bridge is to be constructed over the East River in New York City. The space between the supports needs to be 1050 feet; the height at the center of the arch needs to be 350 feet. Two structural possibilities exist: the support could be in the shape of a parabola or the support could be in the shape of a semi ellipse. An empty tanker needs a 280 foot clearance to pass beneath the bridge. The width of the channel for each of the two plans must be determined to verify that the tanker can pass through the bridge.


A. Determine the equation of a parabola with these characteristics.
B. Determine the equation of a semi ellipse with these characteristics.

[colbtech]

I have no idea how to do this, but as I enjoy this sort of problem. This is how I would tackle it....

The strongest bridge I would presume is based on a circle (?), so....

Take the 1050' as a chord (A-B) in a circle and 350' is the depth of that chord (C-D), where C is the mid-point of the chord (A-B).

From this we can now draw a line from (D-A) and calculate the angle (x) made by CDA. In this instance TAN(x) = 525/350, which gives 56.31°, and as we have a triangle and all the angles add up to 180°, the angle made by CAD is 180 – 90(CDA) – 56.31(DAC) = 33.69°.

From this we can calculate the length of the line between D-A, which comes out at 630.97’.

Now if we extend the depth of the chord line (D-C) through point C, and from A at an angle of 56.31° draw a line such that the extended (C-D) line and this new line intersect. This is point Y, which is the centre point of the circle.

Now…draw a line from Y through the centre point (M) of line A-D. Given that we know the following, D-M = 315.485’, angle CDA = 56.31°, we can now calculate the radius of the curve of the bridge using Cos 56.31° = 315.485/x, which = 568.75’.

Now we know that, we need at least 280’ clearance, so…..another triangle

Take our line from Y to D, subtract 70 (350’ – 280’ clearance) new point Z.

So Y – Z is 568.75 – 70 = 498.75. This is one side of the triangle, the hypotenuse is still 568.75. So using:

Cos 498.75 / 568.75 = 0.877
= 28.73°

Now we have the angle we use:

Sin 28.73° = x / 568.75

Muck it about gives x = Sin28.73° x 568.75
= 273.36’

So now we know that 273.36’ either side of centre and the tanker can pass.

A few things not taken into account, the width of the funnel, the tides, wind pressure, and probably a few others as well. If the boat crashes into the bridge, it ain’t my fault! And before everyone and his dog jumps on the bandwagon, sorry I didn't really answer your question about an equation.

Now if any of the resident experts can do the maths and explain it to my rapidly degenerating brain, I would love to know if I am even on the right track.

[ebaines]

Colbtech assumes the shape of the arch is a circular arc, but the question asked for the case of a semi-ellipse and a parabola, not a circular arc. Fortunately this is easier to do than what colbtech proposed!

First the parabola - you have three points that you know the parabola must meet, namely the base at the left and right supports and the max height at the center. So you know the parabola passes through the points:

(-525,0), (0,350) and (525,0)

Hint - it is usually best in these types of problems to take advantage of symmetry, so by choosing the origin as the mid span of the bridge at water level it will make the math easier.

The general equation for a parabola whose vertex is up is:



In this case is 0, since the parabola is symmetric about the y axis. At x = 0 we know y = 350, so that's the value of . So far we have:


Now at x = 525 we know y = 0, so you can solve for A from:



From this you can solve for A. And now to figure out how wide the ship channel is with 280 foot clearance simply find the values of +/- x where y = 280.

Now for the ellipse. Here the general equation is:



Because this is a semi-elipse with center point at (0,0) you know that and are both 0. You can solve for A and B from the boundary conditions, and follow the same approach as for the parabola to find the width of the ship channel.

[colbtech]

Thanks be... clearly my bridge has collapsed :(... on the tanker :eek:, which has resulted in a massive oil-spill. The environmental and logistical problems seem to compound each other as rescue services try to sort it all out.

The New York Times leads with... "Who the hell came up with a bridge based on a semi-circle?"