Deceleration time for studded snow tires

[Heidi Henkel]

I have data from road tests done in someone else's experiment that show that a car with studded snow tires will take on average 23 meters to stop from 25 mph on packed powder at 0 degrees C. From that information, can I figure out what length of time it takes for the car to stop? Can I figure out lengths of time of segments of this process? For example, can I figure out what length of time it takes to travel the first 6 meters of my skid, or what length of time it takes to slow down from 25 to 20 mph, or both? If a car with these tires starts at 25 mph, skids for 6 meters, and hits a guardrail, for what length of time did they skid? (How soon after slamming on the brakes, does this car crash?) Can I get that kind of information, from the stopping distance? Can I use d=.5at^2+vt and then use the Quadratic Equation? (v=initial velocity in the equation) Is this equation adequate for something that doesn't have to do with acceleration due to gravity? And how would I account for the fact that we are not ending with the velocity being zero? If I use that formula, what would I use for a? The initial experiment didn't weigh the cars. Maybe there is a better way to go about it.

 

[Heidi Henkel]

I am using 20m instead of 23 just to learn how to do the math because the numbers are easier. I made a chart of 1-meter interval of travel, average speed in that interval, speed loss during that interval (average deceleration per meter for that interval), and time elapsed. So now I have 20 different increments. Now let's say I want to find elapsed time over a 6-meter interval. I could add up the six average velocity values for the six one-meter increments, divide by 6, to get average velocity over the 6-meter interval, then multiply the reciprocal by 6 to get elapsed time. Well, what really is the most accurate way to get the average velocity during an interval longer than a meter? I could subtract initial velocity minus final velocity (at each end of the 6-meter interval) and then divide that in half and add it to the final velocity to get the midpoint velocity. I could take the midpoint distance and calculate the velocity at that midpoint distance. I could take the midpoint of average velocities over intervals. I wonder which is most accurate. One thing I notice is that no matter how I find the average, the elapsed time ends up being about the same anyway. I am wondering what is the most accurate way to do it. And maybe there is even a formula that makes the chart-making unnecessary?

 

[BvU]

I am using 20m instead of 23 just to learn how to do the math because the numbers are easier

Changing something for such a reason doesn't make sense to me. You have (made or copied) actual measurements and can set up a hypothesis for a relationship and then test it using the observations.
What, exactly, is the measured data ? Can you make a reasonable assumption from what you see on the chart (or did you make up the chart from some assumed relationship?) ?

 

[Heidi Henkel]

They did lots and lots of tests of people stopping with various kinds of tires on various surfaces with various vehicles. I am wondering how to make meaning of it in terms of time until impact if you crash into something. I am oversimplifying it for the purpose of the forum. Here is the study I am using. http://www.wsdot.wa.gov/research/reports/fullreports/551.1.pdf I am not really using the full scope of the study.
One thing I realize is that it's v^2 that varies linearly with d, so to get an average velocity with which to calculate elapsed time, I would need to get a midpoint of v^2. Or I could get a midpoint d and its v^2. Or for that matter a midpoint d and its v. That would be the average v for the interval, then use the reciprocal, times distance in meters, that is the elapsed time. That's cumbersome but I'm pretty sure it's right.

 

[BvU]

I am not really using the full scope of the study

I can believe that. What table(s) are you using ?
Pity this isn't a homework forum (is it homework?) because there the template has

Homework Equations

.

So, to get some focus:

What is the goal/purpose of your undertaking?
What tools have you got available to help you ? ((fully familiar with SUVAT equations?)
How much detail is required ?​

 

[Heidi Henkel]

I am familiar with the equations. Accuracy within .2 seconds is fine.
My table is : 1-meter interval, average speed in that interval (v for the d halfway through the interval), elapsed time.
If car is going 25 mph which is 11.176 m/s, and has a stopping distance of 20 m due to road surface and tires, then from 20 meters to go to 19 meters to go (starting at 11.76 m/s), Vave is 11.0354 because that's V at 19.5 meters to go using V1^2/V2^2=d1/d2. The elapsed time is (1/Vave)(d) which is 0.906. I made the Vave formula into Vave= square root of (6.245d) with d meaning the midpoint (19.5 in this first interval). Then I can quickly calculate each Vave without going through the whole formula.

So I can make a chart like that. I did 3 intervals then I decided to do it another way.

I am trying to figure out how long a time, approximately, a car spent skidding at me before it hit me, and why I felt like it happened very quickly.

The skid marks were 28 ft long from a car with a 9 ft wheel base, so the car skidded 19 ft which is 5.8m. It was going between 15 and 35 mph. I don't want to get the other person in trouble but I want to make a realistic model of the length of time being possible quite short, so I will try, what does it look like at 20 mph?

I can just use that formula I just mentioned, using 11.176 m/s and 20m for V1 and d1. Then for V2 I use 20mph which is 8.94m/s. For d I get 12.8m. (I could do it leaving the speeds in miles per hour, 25 and 20, I get the same answer.)

So at 20 mph the car has a stopping distance of 12.8m. In a 5.8m skid, it goes from 12.8 m to go, to 7m to go. At 7m to go until stopping, it hits me. At 7m to go, the speed of the car is14.8 mph. Vave over the interval is V at 9.9m to go, which is 17.6mph which is 7.868 m/s.

The elapsed time is (1/7.868)(5.8) = 0.737 seconds. That is not a very long time.

So, you see, in that road condition (packed powder), that car could have slammed on their brakes at 20mph and 28ft from me, and skidded for 0.737 seconds and then hit me. That is a very short time, a very long-looking skid mark, and yet not a very fast speed.

 

[Heidi Henkel]

How I am using the data from that study is just to get a realistic ballpark figure of a stopping distance on that road surface with studded snow tires. I don't have a possibility of being more accurate than a ballpark figure, but the ballpark figure is very helpful as opposed to having no idea what order of magnitude the stopping distance would be. The stopping distances vary in that study, but they are of a consistent range and order of magnitude, which is good enough.

It is not homework.

 

[BvU]

The simplest model for your stopping distances is that by slamming the brakes, thus locking the wheels, the car goes into uniformly decelerated linear motion. The deceleration (negative acceleration) is braking force divided by car mass. The braking force can be expressed as friction coefficient times normal force. Normal force is weight of car = mass of car times gravitational acceleration. In other words, deceleration = friction coefficient times gravitational acceleration. Friction coefficient depends on tyres (studs, all season, blizzaks) and road conditions (icy, snowy, wet, etc.). In formulas:

$d\quad$distance (as function of time)
$d_s\quad$stopping distance
$v_0\quad$ initial speed (11 m/s $\approx$ 25 mph)
$a\quad$acceleration (a<0)
$t\quad$time
$t_s\quad$stopping time
$g\quad$gravitational acceleration (9.81 m/s²)
$\mu_k\quad$ friction coefficient (kinetic -- valid for the duration of the motion)
$m$ car mass; weight is $mg$
$F_f\quad$ friction force
$F_N\quad$ Normal force. $F_N = mg$
"postitive = to the right, negative = to the left "

uniformly decelerated means you assume $\mu_k$ is constant.​

Relations:
$$d=v_0 t + {1\over 2} a t^2 \quad (1)\\ \mathstrut \ \\
v = v_0+at\ \Rightarrow t = {v-v_0\over a} \quad (2) \\ \mathstrut\ \\
(1)\&(2)\Rightarrow\ d = v_0 {v-v_0\over a} + {1\over 2} \left ({v-v_0\over a} \right ) ^2 = {v^2-v_0^2\over 2 a}
$$This last one you already had at hand. Also:$$
t_s = -{v_0\over a}\\
d_s = -{v_0^2\over 2 a}\\
F_{f, {\rm max}} = -\mu_k F_N = -\mu_k mg\\
F_{f, {\rm max}} = ma\\
a = - \mu_k g
$$

 

[BvU]

My table is : 1-meter interval, average speed in that interval (v for the d halfway through the interval), elapsed time

can't find it

 

[BvU]

You have a circumloquacious way of asking something like

I was hit sideways by a car that claims to have been driving between 15 and 35 mph. The skid marks were 20 ft and it covered that distance in what looked like less than a second. Was it really driving that slow or was it driving a lot faster ?
​

What is the actual question ?

 

[Heidi Henkel]

can't find it

I made one. It's not in the other person's study.

 

[Heidi Henkel]

I am just trying to calculate the possible kinematics of a situation in which I was hit by a car that skidded 5.8m before it hit me. So I can calculate it for a variety of possible speeds of the car. It was studded tires on packed snow. The dynamics that I experienced could have occurred if it was going 20 mph. There is no way to know how fast it was going, because some of the energy went into damaging both cars, and there is no way to measure that. I am not looking to calculate the speed of the other car. I am looking to be able to illustrate the basic dynamics numerically.

The first thing you wrote is where I started. Then I was told to do it with V1^2/V2^2 = d1/d2. I can get the weight of the car by looking up the specs of that type of car online. However, I do not have a coefficient of friction for studded tires on packed powder road surface. Maybe I could find that but I have looked quite a bit and have not found such a thing. What I do have is data on stopping distances for cars with studded tires on this and various similar road surfaces. I can use that data to find a way to describe numerically the dynamics of the accident. I don't need a coefficient of friction if I have data on stopping distances. Is there anything wrong with the way I did it? I suppose I could use their data to create a coefficient of friction, but it seems like it's the long way around the barn?

 

[Heidi Henkel]

One thing I am wondering is, if I hit the brakes, what is this uniform deceleration relative to? Distance? Time? I thought v^2 was linear relative to distance because kinetic energy dissipates linearly relative to distance and kinetic energy = .5mv^2. Is one way wrong and the other right? Or are they both right?

 

[Heidi Henkel]

I do not actually know how I would get a coefficient of friction from that study because they did not measure the weights of the vehicles. One thing that is nice about that data is this accident took place right around freezing, so their data around freezing is interesting.

 

[Heidi Henkel]

Another thing that is interesting about this accident is that the cars spun. Some of the energy went into spinning the cars. Not really far. Less than 90 degrees. It's interesting because the cars spun on an axis of their centers of mass. I suppose if I want to get really fancy I could estimate the angle of spin and figure out how much energy it took to spin them, but it's not a high priority to me. I also don't know the weight of my vehicle because it had a lot of stuff in it.

It is important to me to model the deceleration accurately. I really like the V^2 being linear relative to distance, seems realistic based on lots of other life experience doing sports and driving and such.

 

[jbriggs444]

One thing I am wondering is, if I hit the brakes, what is this uniform deceleration relative to? Distance? Time?

Constant acceleration is constant. If you graph it as a function of time, it is constant. If you graph it as a function of distance, it is still constant. If acceleration were variable then its graph as a function of time and as a function of distance would have different shapes. But constant functions are horizontal lines always.

 

[Heidi Henkel]

Then what's up with the v^2 and d ratio equation? This does not produce a constant acceleration.

 

[jbriggs444]

Then what's up with the v^2 and d ratio equation? This does not produce a constant acceleration.

Why ever not?

In fact, v^2 being a linear function of d implies constant acceleration.

 

[Heidi Henkel]

so if I have a car going 25mph, or 11.176m/s, that takes 20m to stop after hitting the brakes, how would I calculate the acceleration rate?

 

[BvU]

See post #8.
$$

d_s = -{v_0^2\over 2 a}\qquad

a = - \mu_k g
$$

 

[Heidi Henkel]

OK so they are both right. I am interested in how I would calculate the constant rate of deceleration. So I can get a value for acceleration which I can use for the familiar quadratic equation for distance vs time. It would be nice to be able to use that formula, but I am confused how to represent a in that equation. d = .5at^2 + Vit + di (the i's are supposed to be subscripts)

 

[Heidi Henkel]

Our messages crossed. Thanks for the formula. a contains coefficient of friction, which I do not know how to calculate based on my data.

 

[jbriggs444]

Our messages crossed. Thanks for the formula. a contains coefficient of friction, which I do not know how to calculate based on my data.

Look at the first formula below. You know distance $d_s$ and starting velocity $v_0$. Solve for acceleration $a$. Now you know acceleration of the car and can look up the acceleration of gravity $g$. You can solve for the coefficient of friction $\mu_k$ if desired.

See post #8.
$$

d_s = -{v_0^2\over 2 a}\qquad

a = - \mu_k g
$$

 

[Heidi Henkel]

20m = -124.9/2a so a = -0.32m/s^2?
I don't have a mu character on my keyboard but it would be 0.03265.

So now I can use the distance quadratic equation, neato! And equations with coefficients of friction.

 

[Heidi Henkel]

This is awesome. One last thing I am curious about. I am not familiar with v^2=v[sub0]^2+2a[delta]x from the SUVAT equations. I am wondering what that is about.

 

[BvU]

Energy: ${1\over 2}mv^2$ is kinetic energy. $ma\Delta x$ is energy (work) done by a force $ma$ over a distance $\Delta x$

 

[Heidi Henkel]

OK this looks rather algebra (distributive property) intensive and possibly not really needed for what I am doing. Fun, but I think I'll give it a rest there, unless you think somehow this will allow me to do something really awesome with the data.

 

[Heidi Henkel]

What is the origin of the d=-v^2/2a equation? Where does that come from?

 

[BvU]

What is the origin of the d=-v^2/2a equation? Where does that come from?

Energy. The kinetic energy from the moving car is dissipated by the friction force the road exerts on the tyres.

 

[Heidi Henkel]

Energy is not part of that equation. Kinetic energy E=.5mv^2. How do you get from that, to the equation that starts with d?

 

[Heidi Henkel]

2ad=v^2
ad=.5v^2

m=1/a? mass is inversely related to acceleration, I believe that...but m=1/a, not necessarily.

 

[OmCheeto]

Energy. The kinetic energy from the moving car is dissipated by the friction force the road exerts on the tyres.

Really? I just rediscovered the "kinematics equations" about a month ago. They are a godsend for idiots like myself.

if $$v_0^2 = 0$$
then the following kinematic equation
$$v^2 = v_0^2 + 2 a \Delta x$$
reduces to
$$v^2 = 2 a \Delta x$$
which rearranged becomes
$$\Delta x = v^2 / 2 a$$

hmmmm... kinetic energy:
$$ke = 1/2 m v^2$$
f = m a --> m = f / a
therefore
$$ke = 1/2 (f/a) v^2 = f v^2 / 2a$$
and
$$W = \Delta ke = fd$$
rearranged yields
f=ke/d
and since ke_f = 0
we get
$$ke = ( ke/d ) v^2 / (2a)$$
multiplying both sides by d/ke yields, once again
$$d = v^2 / 2a$$

Ha! That works too. Thanks!

[edit: Why do I always catch my errors after pushing the post button...]

 

[OmCheeto]

Energy is not part of that equation.

That confused the heck out of me too! But I know from experience, that they don't give out Homework Helper badges to just anyone.

 

[BvU]

It's not really rocket science... (which itself is rather overrated in my opinion ) and a badge (or a PhD, for that matter) is no guarantee against stupidities.

You can check wikipedia or hyperphysics for the formulas.

Interesting reading, this 2002 pdf. I notice Table 5 and table 6 show big differerences in deceleration/acceleration.
Table 5:
If we look at the snow ones, the deceleration is around $${ (11 {\rm\ m/s}) ^2\over 2 \times (20 {\rm \ m})} \approx 3 {\rm \ m/s^2} $$ using this $$ {1\over 2} mv_{\rm end}^2 - {1\over 2} mv_0^2 = F_{\rm friction} \times d_s = - ma\times d_s$$ energy dissipation equation . That is pretty good !
The bare pavement ones are plain excellent : $\approx 12 {\rm \ m/s^2}$ ! Hurray for ABS (you did notice that in the text ?)

(question: the 20 feet skid marks in the snow - were they from a vehicle with ABS ? What tyres did it have ?)

Table 6:
Even easier to calculate acceleration: a mere $\approx 1.2 {\rm \ m/s^2}$ ! No traction control, apparently ? (the opposite of ABS).

--

I am puzzled about table 12: much more deceleration, and no explanation in the text. Makes one suspect that friction is better at higer speeds...

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