How would I calculate the centripetal acceleration of a car on an elliptical path?

[JimEd]

How would I calculate the centripetal acceleration of a car at any (every) point on an elliptical path?

Assume the car's speedometer is showing a constant speed the entire time.

Answer in terms of g forces, please. (And, I have never taken calculus).

You could also assume zero g at the flattest part of the ellipse, and 1g at the tightest part of the ellipse, if that helps.

Thanks.

 

[Meir Achuz]

You have to use the 'radius of curvature' for where you are on the ellipse. You have to a bit of calculus in order to find the radius of curvature, although some book might have the formula for an ellipse. There is never zero g, even at the flattest part of an ellipse.

 

[zhangyang]

You can use the parameter equation of a motion along an elliptical path:x=x(t),y=y(t).Derive them twice ,and get the acceleration toward x-axis and y axis.
Second ,derive the common equation y=y(x),find the tengent,than you can get the vertical direction,it is also the centripetal direction.
Third,decompose the two acceleration to the centripetal direction,and add them.So,the centripetal acceleration is worked out.

 

[JimEd]

You have to use the 'radius of curvature' for where you are on the ellipse. You have to a bit of calculus in order to find the radius of curvature, although some book might have the formula for an ellipse. There is never zero g, even at the flattest part of an ellipse.

Hi Clem,

Could you show me an example? Good point about zero g.

Thanks

 

[JimEd]

You can use the parameter equation of a motion along an elliptical path:x=x(t),y=y(t).Derive them twice ,and get the acceleration toward x-axis and y axis.
Second ,derive the common equation y=y(x),find the tengent,than you can get the vertical direction,it is also the centripetal direction.
Third,decompose the two acceleration to the centripetal direction,and add them.So,the centripetal acceleration is worked out.

Hi Zhangyang,

Could you show me an example? In steps, if you have the time.

Thank you.

 

[JimEd]

Anyone else? Help?

 

[Ken G]

If you have motion at constant speed, then any acceleration you have must be centripetal acceleration. It will point perpendicular to the instantaneous direction of motion, as if the tiny piece of curve the particle is on is a tiny piece of a circle. At some point you will have to have the equation of an ellipse, so you can calculate the acceleration. If you can get the radius of curvature r at any point, you know the acceleration has magnitude v²/r, where v is the constant speed. If you instead have the y(x) of an ellipse, you are going to need calculus to figure out the acceleration subject to a constant v. Calculus tells you that v_y=dy/dt=dy/dx*dx/dt=dy/dx*v_x, and you know that (v_x² + v_y²)^1/2 = v. Substitute and you get that v_x/v = [1 + (dy/dx)²]^1/2 and v_y/v = dy/dx*[1 + (dy/dx)²]^1/2, where you know dy/dx at any x from you ellipse equation. Think of these as two functions of x, call them X(x) and Y(x). The acceleration is the vector whose x and y components are the d/dt of those two functions, acceleration = (v*dX/dt,v*dY/dt), where dX/dt = dX/dx*X*v and dY/dt = dY/dx*X*v. That gives you acceleration as a function of x.

 

[zhangyang]

Hi Zhangyang,

Could you show me an example? In steps, if you have the time.

Thank you.

I have made a mistake,I have not noticed that the volecity is a constant.
If so,I will use the geometrical way:the bending of a curve.This problem can not avoid calculus.The bending of a curve is defined as:

ds/dθ=1/r.

A very short part of an elliptical path can be treated as a circular arc.You can find the centre of the circular arc.
ds denote the length of arc,dθ denote the angle. r denote the radius of the arc.
And use the formula :

centripetal acceleration =v/r.

But the r is calculated by a calculus way. So calculus is absolutely necessary in this problem,you can find a textbook,and find the geometrical meaning of calculus,such as derivative and tangent.

 

[zhangyang]

centripetal acceleration =v*v/r.

 

[rcgldr]

If you can get the radius of curvature r at any point.

If the function for the ellipse is a function of y(x), where any point on a curve is defined as {x, y(x)}, then radius of curvature at x is:

radius(x) = (1 + y'(x)²)^3/2 / | y''(x) |

where y' is first derivative of y, and y'' is second derivative of y. Since speed is constant in this case, then acceleration at x = v² / radius(x).

Not used here, but if the function is defined by {x(t), y(t)}, then

radius(t) = ( (x'(t))² + (y'(t))² )^3/2 / |(x'(t) y''(t)) - (y'(t) x''(t)) |

Then again, if the function is defined by {x(t), y(t)}, then you can directly calculate the total acceleration, even if the path isn't curved or acceleration isn't always perpendicular to velocity as it would be in a constant speed case:

acceleration(t) = (x''(t)² + y''(t)²)^1/2

 

[JimEd]

Wow, great answers. Thank you all!

You have given me much to think about, and try to understand. But if anyone has anything to add, please feel free.

Thank you again.

 

[Philip Wood]

I found this formula on a website...

R = [itex]\frac{[(bx)² - (ax)² + a⁴]^{$\frac{3}{2}$}}{ba⁴}[/itex]

Apologies for latex problem: don't know what's gone wrong, but it's easy to interpret.

The formula gives the radius of curvature at x = x for the ellipse
[itex]\frac{x²}{a²}[/itex] + [itex]\frac{y²}{b²}[/itex] = 1.

The magnitude of the acceleration is, then, v²/R.

 

[JimEd]

Thanks Philip, that looks great, what website if you don't mind me asking?

And thanks, Doc Al.

 

[Doc Al]

I found this formula on a website...

R = [itex]\frac{[(bx)² - (ax)² + a⁴]^{$\frac{3}{2}$}}{ba⁴}[/itex]

Apologies for latex problem: don't know what's gone wrong, but it's easy to interpret.

You're mixing latex and non-latex commands. It's easier to stick to latex:

$$R = \frac{[(bx)^2 - (ax)^2 + a^4]^{\frac{3}{2}}}{ba^4}$$

 

[Philip Wood]

Doc Al. Thanks. The mixing wasn't intentional. I used the latex reference on this site. Are the commands – those which are unintentionally visible in my post – non-latex, then?

Jim Ed. I'm worried that there might be a PF rule against giving URLs, but I typed 'curvature of ellipse' into a well known search engine, and found the formula under the title 'Old notes on curvature'.

The formula is dimensionally alright, gives neat values of R when x = a, and when a = b (case of circle).

 

[Doc Al]

Doc Al. Thanks. The mixing wasn't intentional. I used the latex reference on this site. Are the commands – those which are unintentionally visible in my post – non-latex, then?

You mixed in things like 'sup' tags (superscript). Like this: x²; in latex it would be: $x^2$.

 

[Philip Wood]

Ah! It all comes clear. I foolishly assumed that the 'sup' tags would produce latex commands, and were all part of the latex toolkit given by PF.

 

[JimEd]

Doc Al. Thanks. The mixing wasn't intentional. I used the latex reference on this site. Are the commands – those which are unintentionally visible in my post – non-latex, then?

Jim Ed. I'm worried that there might be a PF rule against giving URLs, but I typed 'curvature of ellipse' into a well known search engine, and found the formula under the title 'Old notes on curvature'.

The formula is dimensionally alright, gives neat values of R when x = a, and when a = b (case of circle).

Thanks for that Philip, I found it, with your help. Any chance I could PM you sometime if I have more questions?

 

[Philip Wood]

JimEd. Don't know what 'PM' means, but I'm flattered anyway! I don't know very much about anything, though. In the case of the curvature, I was going to work it out by the general formula in terms of derivatives, but then thought: why bother – it's bound to be on some site or the other!

 

[JimEd]

JimEd. Don't know what 'PM' means, but I'm flattered anyway! I don't know very much about anything, though. In the case of the curvature, I was going to work it out by the general formula in terms of derivatives, but then thought: why bother – it's bound to be on some site or the other!

Oh, PM is Private Message (look in the upper right hand corner of this page, under where it says: "Welcome, Philip Wood"). It's kind of like email. It sure looks like you know your stuff to me!