# Path of a car accelerating around a corner

If a car is going around a 360 degree circle at 1 g (its maximum centripetal acceleration, and its maximum total acceleration) and then longitudinally accelerates ("gives it the gas") at .5 g (or tries to, while still cornering) for the next 45 degrees, what would be the formula for the radius the car is on while it is longitudinally accelerating?

How would it's velocity be calculated at points along its path?

(You can assume a velocity and a radius for the initial 360 degree circle if you need to. You can assume the centripetal acceleration decreases at the same rate the longitudinal acceleration increases if you need to.)

Thanks.

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tiny-tim
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Hi JimEd! Show us what you've tried, and where you're stuck, and then we'll know how to help! Hi, thanks.

I don't know much... a=v^2/r, Total g = sqr (longitudinal g)^2+(lateral g)^2, v=sqr (g*r)... I don't really even have a plan of attack on it.

I'm an older guy, so this isn't homework (I'm not against learning, I'm just letting you know I'm not cheating on my homework!), this is just something I've been wondering about.

rcgldr
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If a car is going around a 360 degree circle at 1 g (its maximum centripetal acceleration, and its maximum total acceleration) and then longitudinally accelerates ("gives it the gas") at .5 g (or tries to, while still cornering).
Assume maximum acceleration is 1g regardless of direction with respect to the car. Then the equation for acceleration is

1 = sqrt(lateral2 + longitudinal2)

in this case

1 = sqrt(lateral2 + .52)
1 = lateral2 + .52
lateral2 = 1 - .52
lateral = sqrt(1 - .52)
lateral = 0.866 g

So acceleration perpendicular to velocity is .866 g, and acceleration in the direction of velocity is .5 g. Also v2 / (radius of curvature of the path) = .866 g. I'm thinking you'll have to solve a differential equation to determine the path at this point, perhaps using polar coordinates, but not sure how to proceed.

Thanks. That was pretty much where I had gotten to with a=v^2/r and Total g = sqr (longitudinal g)^2+(lateral g)^2, so it does verify I am on the right track.

Can anyone help me with what the radius would be?

Thanks, but that is way over my head.

I was thinking there might be some ratio/equivalency of 1=v^2/r and .866=v^2/r to give the new radius...

rcgldr
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Since acceleration in the direction of velocity is .5 g, I think this means

speed = speed0 + . 5 g t

If you define speed0 = 0 at t = 0:

speed = . 5 g t

However all this results in is

radius of curvature = (. 5 g t)2 / .866 g = 0.288675 g t

Since radius of curvature is a combination of first and second derivatives, this doesn't help.

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So... the rate of change of the radius is .288675 (meters/degree?)

rcgldr
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So... the rate of change of the radius is .288675 (meters/degree?)
No, that's the rate of change of the radius of curvature versus time, = .288675 g t.

The alternative would be to use numerical integration like Runge-Kutta with the acceleration values to create a set of points and generate a graph.

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At t =0, r should be = a/v^2 since the two radii (initial 360 degree circle and accelerating arc) are tangent at that time, correct?

Would that help?

rcgldr
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update - It's the radius of curvature that is = v2 / a, and not the polar coordinate r. I was using r to mean radius of curvature which was confusing, so I corrected my previous posts.

It would be easier to just use a spiral for the entire path, instead of transitioning into a spiral from circle at some initial point r and θ or x and y. This could elimnate constants of integration related to the initial condition at t = 0.

I'm not sure if cartesian or polar coordinates would be better in trying to solve this problem directly instead of using numerical integration.

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Do we have any idea what the formula would be for the spiral?

Could it be an elliptical segment instead?

Finding the shape is really what I would like to know.

rcgldr
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Do we have any idea what the formula would be for the spiral?
I don't, at least not yet. Only that acceleration is a 1g vector, 60° "inwards" inwards of the current velocity vector. I'll try to setup numerical integration on a spread sheet later on to get an idea.

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Thanks rcgldr. I really do appreciate your help.

I was wondering if there might be some "boost a spacecraft to a higher orbit" formula we could use.

Except that gravity decreases with higher orbits...

rcgldr
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correction to previous post:

radius of curvature = (. 5 g t)2 / .866 g = 0.288675 g t2

I used numerical intergration to get a path. The forward acceleration increases velocity quick enough that the path ends up looking a bit like a parabola. Link to graph image:

http://rcgldr.net/misc/carpath.jpg

To get a spiral, the forward acceleration would need to be much smaller, like .1 g:

http://rcgldr.net/misc/carpath0.jpg

The equation I use starts with the forward g component of acceleration Af (I used .5 and .1), to get the angle of acceleration versus velocity vector:

θ = cos-1(Af)

acceleration vector equals velocity vector rotated θ (counter-clockwise), with magnitude changed to 1g:

Ax = g (cos(θ) Vx - sin(θ) Vy) / sqrt(Vx2 + Vy2)
Ay = g (sin(θ) Vx + cos(θ) Vy) / sqrt(Vx2 + Vy2)

Numerical integration was used to generate velocities and positions for the graph images I made.

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That first graph looks perfect. Thank you very much! Now I have tons of questions! (But you have certainly done more than your share if you have had enough of this and would like to leave it to someone else).

What are the numbers/units on the scale of the graphs? To form a physical path they would have to be distances/lengths, I'm guessing.

How would:
θ = cos-1(Af)
Ax = g (cos(θ) Vx - sin(θ) Vy) / sqrt(Vx2 + Vy2)
Ay = g (sin(θ) Vx + cos(θ) Vy) / sqrt(Vx2 + Vy2)
be written differently if lateral g = something other than 1 (say, 2 g)?

When you say θ = cos-1(.5) = 60 is the angle of acceleration versus velocity vector... and you had previously said that acceleration is a 1g vector, 60° "inwards" of the current velocity vector... Basically, you are saying that the combination of the two accelerations: .866 g inward/laterally, and .5 g forward/longitudinally result in a 1 g acceleration 60 degrees (mostly inward/laterally) from the current direction of travel of the car. Is that right?

Then, "acceleration vector equal to the velocity vector rotated θ (counter-clockwise)". That is another way of stating what we just said in the paragraph above, right? And "Counter-clockwise" would be the same as "inward".

After you have that angle, the formulae for Ax and Ay give the current x,y proportions of the total acceleration. The first Vx and Vy to plug into these formulae would come from the velocity of the original 360 degree circle? And then change each time after that?

I don't know anything about numerical integration (as you've probably surmised), but is there any way I could draw the original circle and then manually plot a few x,y coordinates and play connect-the-dots to make a rough version of my own graph?

Thank you again for all you have done, and walking me through this.

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rcgldr
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What are the numbers/units on the scale of the graphs?
Meters, and I used 1 g = 9.80665 m / s2.

How would:
θ = cos-1(Af)
Ax = g (cos(θ) Vx - sin(θ) Vy) / sqrt(Vx2 + Vy2)
Ay = g (sin(θ) Vx + cos(θ) Vy) / sqrt(Vx2 + Vy2)
be written differently if lateral g = something other than 1, say, 2 g?
Just multiply Ax and Ay by 2:

Ax = 2g (cos(θ) Vx - sin(θ) Vy) / sqrt(Vx2 + Vy2)
Ay = 2g (sin(θ) Vx + cos(θ) Vy) / sqrt(Vx2 + Vy2)

When you say θ = cos-1(.5) = 60° is the angle of acceleration versus velocity vector... and you had previously said that acceleration is a 1g vector, 60° "inwards" of the current velocity vector... Basically, you are saying that the combination of the two accelerations: .866 g inward/laterally, and .5 g forward/longitudinally result in a 1 g acceleration 60° (mostly inward/laterally) from the current direction of travel of the car. Is that right?
Yes, sin(60°) = .866, cos(60°) = .5. Note 60° = π/3 if you want the angle in radians.

Then, "acceleration vector equal to the velocity vector rotated θ (counter-clockwise)". That is another way of stating what we just said in the paragraph above, right? And "Counter-clockwise" would be the same as "inward".
Lateral accleration can be "inwards" if it's clockwise or counter-clockwise. For clockwise rotation, θ would be negative, but the acceleration would still be "inwards".

The first Vx and Vy to plug into these formulae would come from the velocity of the original 360 degree circle?
The choice for the initial {x, y}, {Vx, Vy} could be anything (as long as velocity isn't zero). I chose to use {x, y} = {1, 0} and {Vx, Vy} = {0, sqrt(g)}, which would correspond to the tangent of a circle with a radius of 1.

And then change each time after that?
The numerical integration does that.

I don't know anything about numerical integration (as you've probably surmised), but is there any way I could draw the original circle and then manually plot a few x,y coordinates and play connect-the-dots to make a rough version of my own graph?
You can download this zip of the .xls files and use a spreadsheet program. Note that once you get a graph, you have to adjust the size to get the axis to scale equally. I literally use a ruler next to my screen to do this with my version of excel.

http://rcgldr.net/misc/carpath.zip

For numercial integration I used RK4:

wiki rk4.htm

to integrate {Ax, Ay} into {Vx, Vy}. On the carpath spreadsheets, {dvx1, dvy1} (which use {ax, ay}), correspond to k1 from the wiki article. {dvx2, and dvy2} correspond to k2, {dvx3, and dvy3} correspond to k3, {dvx4, and dvy4} correspond to k4. k1, k2, k3, and k4 are ΔV's (change in velocities based on Δt times a weighted average of accelerations), so {Vx, Vy}: Vi = Vi-1+(k1+2 k2+2 k3+k4)/6.

Since position isn't a feedback into acceleration, to integrate position {x, y} : Pi, I used average velocity

Pi = Pi-1 + Δt (Vi-1+Vi+Vi+1)/3

To try different values, all you need to do is adjust the angle, g force and time step which are at the top of the spreadsheet (h2, i2, j2) and/or copy paste more rows if needed.

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Thanks for that. A lot to think about!

I'm sure I will have more questions after I have had time to cogitate.

Thank you again. rcgldr. I think I am starting to understand this, thanks to you. Is there a place on this board where I can give you a gold star? You surely deserve one.