-1
$\begingroup$

I have a car traveling on an elliptical race track at a constant angular velocity of A radians/sec. The angular velocity is calculated at the intersection of semi-major & semi-minor axis. I know the eccentricity e, semi-major axis a, semi-minor axis b of the ellipse & the time T it takes for completing one lap of the race track.

I am not much familiar with calculus related to elliptical shapes. For a circle, the distance traveled can be found through Speed-Distance-Time formula. But for an ellipse, to maintain constant angular velocity, the linear velocity needs to be changed continuously. The car needs to go slower around semi-minor axis point & faster around semi-major axis point as far as I know.

I tried to find how the linear velocity should change but couldn't make sense of it. I also tried to approximate the distance measurement using the general formula
$$ \frac{Lap \ Circumference}{Lap \ Period} = \frac{Distance \ Travelled }{Time \ to \ travel\ Distance \ length}$$
I am looking for a formula to calculate this.

How to find the actual distance in meters traveled by the car after time t. Assume that the start/finish line is at semi-major axis point.

$\endgroup$
11
  • $\begingroup$ How familiar are you with elliptic integrals? $\endgroup$ – Henry Jun 21 '18 at 9:59
  • 1
    $\begingroup$ Elliptic integrals are not elementary calculus, though numerical approximations are possible $\endgroup$ – Henry Jun 21 '18 at 11:10
  • 1
    $\begingroup$ Is the angular velocity measured about the common mid point of the axes or...? Say, about one of the focal points of the ellipse? Given your fascination with observational astronomical data the latter sounds more natural actually :-) $\endgroup$ – Jyrki Lahtonen Jun 22 '18 at 14:33
  • 1
    $\begingroup$ @JyrkiLahtonen The angular velocity in an ellipse will be different at different places as it is measured at the intersection of semi-major & semi-minor axis. The value of angular velocity I have is an average value for one complete lap. $\endgroup$ – KharoBangdo Jun 23 '18 at 4:38
  • 1
    $\begingroup$ Ok. In that case all you need to do is to calculate the arclengths as in the linked question. If I got it right for your purposes it suffices to cover integer multiples of quarter laps. Unfortunately there is no simple closed formula for the length of a quarter lap around an ellipse. Elliptic integrals are a pain :-/ $\endgroup$ – Jyrki Lahtonen Jun 23 '18 at 6:23
0
$\begingroup$

HINTS:

Since $ \theta = \omega t $ where $\omega$ is constant we can do the integration w.r.t. $\theta$ or time. In central ellipse polar form

$$ \frac{\cos^2 \theta }{a^2}+\frac{\cos^2 \theta }{b^2}= \frac{1}{r^2} \tag1 $$

Differentiate w.r.t. arc ( primed w.r.t. arc $s$ )

$$ {\theta^{'}} \,\sin 2 \theta (1/b^2-1/a^2) =\frac{-2{r^{'}} }{r^3} \tag2 $$

Distance along arc can be further found by evaluating through radius $r$

$$ ds^2=dr^2 + (r\, d \theta)^2; \quad \tag 3 $$

$${r{'} ^2 + {({ r\theta^{'}} )^{2}} } = 1 \tag4 $$

Solve for $ ( r^{'}, \theta^{'}) $ from (2) and (3) to form two coupled differential equations and solve for $r, \theta= \omega t. $

$\endgroup$
8
  • $\begingroup$ How did you go from 2nd equation to 3rd equation? $\endgroup$ – KharoBangdo Jun 25 '18 at 9:54
  • $\begingroup$ It is independent. A Diffrl Equn should be formed as indicated $\endgroup$ – Narasimham Jun 25 '18 at 19:08
  • $\begingroup$ I must mention that I've forgotten many mathematical concepts which aren't mostly used in my field (DSP). I know Eq (1) of the ellipse & I know Eq (2) is a derivative of (1) wrt r. Now 'r' is the distance traveled on the elliptical path, '$ \theta % is the corresponding angle made. So what is 's'. Also I am not familiar with Eq (3.1). It looks like Pythagoras equation. If so, how did you make a right angled triangle. $\endgroup$ – KharoBangdo Jun 26 '18 at 4:57
  • $\begingroup$ $r^2=x^2+y^2;$ Arc $s$ is distance traveled along ellipse arc; $(r,\theta)$ are polar coordinates; Eq 3.1 is differential form of Pythagoras.... If $(r,s)$ are given for ellipse with focus as origin can you convert them to the central polar form ? $\endgroup$ – Narasimham Jun 26 '18 at 8:05
  • $\begingroup$ After reading Pythagorean theorem, I understood Eq 3.1 & 3.2. Then you wan't me to substitute r' from Eq 3.2 into Eq. 2 so that I have diff eq wrt ds & d(theta). But upon substituting, I still have r which is unknown right? Is that correct $\endgroup$ – KharoBangdo Jun 26 '18 at 8:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.