I'm taking electronics technology at a college and my calculus teacher gave us a civil engineering assignment where we have to design a highway turn. A straight line to a spiral to an arc to a spiral to a straight line. I have to find the polynomial equation of my spiral and my calculus teacher is friggin scattered.

All I have to work with is a radius of the arc which is 435.468 meters. Also my overall angle is 63 degrees (10 for each spiral chunk and 43 for the arc). The velocity is 100 km/h, the acceleration during the turn is 1.7719 m/s^2 and my time spent in each spiral is 2.7361 s.

The general equation for the spiral is y=a+bx+cx^2+dx^3+ex^4+fx^5. x is the unknown and a to e are constants. All I know is that my derivative from the straight line to the spiral is 0 and the end of the spiral will have the same tangent as the beginning of my arc.

Anyone want to explain to me how the hell I go about solving this? If I havn't provided enough info just ask.


BTW, if anyone is wondering why I'm doing such an assignment is because people who are taking chemical, mechanical, and biomedical technology have to do the same thing and civil engineering was neutral ground (course not available) so no one could bitch about it. The teacher was too lazy to make individual projects I guess.

A straight line to a spiral to an arc to a spiral to a straight line.

Draw it

I suck at ascii art and have no pics.

Imagine a straight horizontal line. The turn will ultimately be 90 degrees so you end up on a vertical. Instead of an edge you have an arc.

Since an arc has a fixed radius, you cant just instantaneously turn your steering wheel from 0 to whatever angle. During the rate of change of the steering wheel you're in a spiral (constant change in radius). Once you hold the steering wheel in a static position, that's the arc. And then a spiral again until you get back on a straight line.

Hope that helped.


Edit: That turn is 63 degrees not 90, sorry.

2nd Edit: I'm just looking for the equation of the first spiral, not the second.

is the acceleration just centripetal (i.e. tangential velocity is constant)? does it matter where you turn off the straight line? i'm guessing you want complete continuity (i.e. where the line becomes the spiral and where the spiral become the arc the derivatives are equal)? if you can use the speed data to calculate the length of the "spiral" that might help, although this problem sounds like it would be infinitely easier in polar coordinates.

The acceleration is centripetal, and each end of one part is the beginning of the next. So the first point on the spiral will have a derivative of zero and the end point will have a derivative equal to the beginning of the arc. And my teacher wants the equation polynomial.

I still can't visualize this, sorry.

This is how I visualise it...

Draw a square ABCD.
Add a line BE, where BE=BD so that the angle CBE = 10 degrees.
Add a line BF, where BF=BD so that the angle EBF = 43 degrees.
Add a line BG, where BG=BD so that the angle FBG = 10 degrees.
Add a square BGHJ, so that angle ABH is 117 degrees.

In reality BD=BG and BE=BF, but BD does not equal BE.
It's pretty close though.

Add an arc from E to F, with B as the centre of the circle.

The problem is to add the curved line from C to E, and F to G.
These lines are spiral in character, but look almost like an arc.

The combined line that is to be described is D-C-E-F-G-H.

Sorry for the geometry, but that's what works for me.

This is how I visualise it...

Draw a square ABCD.
Add a line BE, where BE=BD so that the angle CBE = 10 degrees.
Add a line BF, where BF=BD so that the angle EBF = 43 degrees.
Add a line BG, where BG=BD so that the angle FBG = 10 degrees.
Add a square BGHJ, so that angle ABH is 117 degrees.

In reality BD=BG and BE=BF, but BD does not equal BE.
It's pretty close though.

Add an arc from E to F, with B as the centre of the circle.

The problem is to add the curved line from C to E, and F to G.
These lines are spiral in character, but look almost like an arc.

The combined line that is to be described is D-C-E-F-G-H.

Sorry for the geometry, but that's what works for me.

Hint:
MSpaint

If I used over 500 words, I might consider a picture.
This doesn't really need one though.

But mash_buttons, these sites could be useful:

http://www.ctre.iastate.edu/educweb/ce353/lec05/lecture.htm
http://mysite.du.edu/~jcalvert/railway/transpir.htm

Cheers.

Ok so I figured it out. At the first point of the spiral (same as a straight line) when X would equal zero would leave just the length of Y which is the radius. So the A coefficient is my radius. Finding y' and y'' makes B and C zero so they disappear entirely from the equation ( I can't go further with third and up derivatives because of two-dimensional limitations). I used basic trig to find X and Y for my second point. ( I'm doing this obviously in rectangular). My first slope is zero and the second one turns out being -X/Y (X^2+Y^2=radius^2, end of spiral is the start of the arc). Since there is three more unknowns (D,E,F), I have to bust out simultaneous equations, which I do by placing my second Y point value equal to the remaining equation and sub the second X point into D,E, and F. In the second equation, I place my second slope equal to y' of the remaining equation and once again place my second X point for D,E, and F. The third equation is the derivative of my second point slope and equate it to y'' of the remaining equation ( sub X again). Then I just solve to find the values of D,E, and F.......It was a bitch though. I end up with Y=radius+(coeff)X+(coeff)x^2+(coeff)x^3.

Sorry if that's hard to follow.