[Date Prev][Date Next][Thread Prev][Thread Next][Date Index][Thread Index]
OT Car crash Physics...
--0-1954395481-1035850515=:33059
Content-Type: text/plain; charset=us-ascii
I used to do accident reconstruction, so maybe I can help.
What do you mean by "back remaind stationary" If the car was hit head on and collapsed 8 feet with the back wheels not moving, you are then left with a jaguar half the original size. Since there is nothing to keep the back wheels in their place, this wouldn't happen anyway. The jag would move a lot before/while collapsing that way.
Did the noble jag pivot about it's rear wheels when struck? Maybe so, if it was hit at more of an angle than your description lead me to believe. The problem is that the rear of the vehicle would never have stayed in place - at least one rear wheel has to move for the car to pivot. How much? At what angle? All of this stuff is figured out by measuring skid marks on the pavement, their exact direction (for each wheel) and mapping the movement of the car onto these marks.
Let me just tell you how we would work an accident reconstruction like this. We would measure the degree to which the jaguar is crushed in and compare that with crash test data on the jag to determine how much energy it takes to crush that car that much. From that you can directly determine the difference in the speed of the two objects with pretty good accuracy. Since you know the jag was stopped, that's all you need.
Another approach would be to measure the displacement of the jag and motorcycle after the crash, and see how much energy they used up in coming to rest. That's more the approach you seem to be going for. There are several catches.
Easiest way is to assume that the car and bike became one after they hit. If that combined mass came to rest after travelling 2.4m, then all you need is some coef of friction and you are set. The problem is again in the crushing of the car. The distance the combined car/bike unit moves has to be taken as the distance the center of mass of the unit moves. If the jag didn't crush much, then it doesn't matter, but if it did, then the center of mass is moving a lot less than the front wheels are.
So if the jag didn't crush much, you can still procede, but what angle was the impact? If the brakes of the jag were fully applied throughout the collision such that the wheels were always locked up, you don't really care much (as long as you have corerectly figured out the displacement), but no one I know holds their brake pedal down to the floorboard when waiting at a light.
If you have gotten through all of this, then you only have to figure out the coeff of friction for a locked up tire on asphalt. There are standard numbers used for this in reconstruction, that I don't have on me (imagine that).
So here's your estimate: if a jag is moved 2.4m by something a tenth its size, then the bike was absolutely hauling ass. Hope that helps.
Jason Adams <roclist@accessconsulting.ca> wrote:M1 is Jaguar XJS curb weight 4808lbs driver 165lbs ---> 2260kg
M2 is Yamaha Motorcycle 360lbs + 125lbs ----->220kg
Jaguar is stationary (waiting to turn left) and is struck by motorcycle
headlong in the pass front fender.
Front of car is displaced 8 feet (2.44m) back remains stationary, wheelbase
is 102" (2.59m)
uK is a coefficient I found on the net.
I want to calculate the inital velocity of the motorcycle.
and no the motorcycle driver isn't doing so well....
Jason Adams
Fahrvergnugen Forever! ;)
84 rocco 16v
93 320i
98 Z71
----- Original Message -----
From: "Jason"
To: "Jason Adams" ; "0sciroccolist"
Sent: Monday, October 28, 2002 1:42 PM
Subject: Re: OT Car crash Physics...
> At 03:57 PM 10/28/2002, Jason Adams wrote:
> >With all the wealth of informed people on the list this shouldn't be too
> >difficult...
> >
> >If I know the weight of the cars, the skid distance, assume complete
> >inelastic collision. how do I work it out...
>
> I'm confused here -- what are you trying to work out?
>
> >Vf^2 = Vo^2 + 2a(dS)
> >
> >is there something wrong with my assumption for 'a' ?
>
> Well, uK isn't a constant... the value you use is a "typical" value.
Every
> tire is different, as is every road surface. And stopping distances on a
> car aren't that easy to compute -- you're dealing with 4 tires on an
object
> that has suspension and that does not have even weight distribution.
>
> "Typical" deceleration for a modern automobile is between 8.5 and 10
> ms^-2. Of course, some are far below, and a few are above.
>
> Using the 3 braking distance figures I have for the 16V from magazines
(60,
> 80mph from Road & Track, 70mph from Car & Driver), we can compute the
> average a for the 16v's braking:
>
> 60mph 150 feet 7.87g
> 70mph 196 feet 8.19g
> 80mph 257 feet 8.16g
>
> The average of those 3 stops is 8.08g. The 8V (158 feet from 60, 271 feet
> from 80), averaged 7.61g.
>
> Of course, modern tires and brake pads improve those distances: Two years
> ago I did 10 consecutive stops from 60 in my16V with Potenza RE71 tires
and
> Ferodo pads and used my G-Tech Pro to measure the distance. I discarded 3
> runs where the G-Tech did not provide an accurate reading, which left me
> with 7 good runs. I discarded the best and the worst runs, leaving 5
> total. The average of those 5 runs was 131.8 feet, which is an average of
> 8.95g.
>
> Jason
>
>
>
>
>
>
>
_______________________________________________
Scirocco-l mailing list
Scirocco-l@scirocco.org
http://neubayern.net/mailman/listinfo/scirocco-l
---------------------------------
Do you Yahoo!?
Y! Web Hosting - Let the expert host your web site
--0-1954395481-1035850515=:33059
Content-Type: text/html; charset=us-ascii
<P>I used to do accident reconstruction, so maybe I can help.
<P>What do you mean by "back remaind stationary" If the car was hit head on and collapsed 8 feet with the back wheels not moving, you are then left with a jaguar half the original size. Since there is nothing to keep the back wheels in their place, this wouldn't happen anyway. The jag would move a lot before/while collapsing that way.
<P>Did the noble jag pivot about it's rear wheels when struck? Maybe so, if it was hit at more of an angle than your description lead me to believe. The problem is that the rear of the vehicle would never have stayed in place - at least one rear wheel has to move for the car to pivot. How much? At what angle? All of this stuff is figured out by measuring skid marks on the pavement, their exact direction (for each wheel) and mapping the movement of the car onto these marks.
<P>Let me just tell you how we would work an accident reconstruction like this. We would measure the degree to which the jaguar is crushed in and compare that with crash test data on the jag to determine how much energy it takes to crush that car that much. From that you can directly determine the difference in the speed of the two objects with pretty good accuracy. Since you know the jag was stopped, that's all you need.
<P>Another approach would be to measure the displacement of the jag and motorcycle after the crash, and see how much energy they used up in coming to rest. That's more the approach you seem to be going for. There are several catches.
<P>Easiest way is to assume that the car and bike became one after they hit. If that combined mass came to rest after travelling 2.4m, then all you need is some coef of friction and you are set. The problem is again in the crushing of the car. The distance the combined car/bike unit moves has to be taken as the distance the center of mass of the unit moves. If the jag didn't crush much, then it doesn't matter, but if it did, then the center of mass is moving a lot less than the front wheels are.
<P>So if the jag didn't crush much, you can still procede, but what angle was the impact? If the brakes of the jag were fully applied throughout the collision such that the wheels were always locked up, you don't really care much (as long as you have corerectly figured out the displacement), but no one I know holds their brake pedal down to the floorboard when waiting at a light.
<P>If you have gotten through all of this, then you only have to figure out the coeff of friction for a locked up tire on asphalt. There are standard numbers used for this in reconstruction, that I don't have on me (imagine that).
<P>So here's your estimate: if a jag is moved 2.4m by something a tenth its size, then the bike was absolutely hauling ass. Hope that helps.
<P> <B><I>Jason Adams <roclist@accessconsulting.ca></I></B> wrote:
<BLOCKQUOTE style="PADDING-LEFT: 5px; MARGIN-LEFT: 5px; BORDER-LEFT: #1010ff 2px solid">M1 is Jaguar XJS curb weight 4808lbs driver 165lbs ---> 2260kg<BR>M2 is Yamaha Motorcycle 360lbs + 125lbs ----->220kg<BR><BR>Jaguar is stationary (waiting to turn left) and is struck by motorcycle<BR>headlong in the pass front fender.<BR><BR>Front of car is displaced 8 feet (2.44m) back remains stationary, wheelbase<BR>is 102" (2.59m)<BR><BR>uK is a coefficient I found on the net.<BR><BR>I want to calculate the inital velocity of the motorcycle.<BR><BR>and no the motorcycle driver isn't doing so well....<BR><BR><BR><BR>Jason Adams<BR>Fahrvergnugen Forever! ;)<BR>84 rocco 16v<BR>93 320i<BR>98 Z71<BR><BR><BR><BR>----- Original Message -----<BR>From: "Jason" <JASON@SCIROCCO.ORG><BR>To: "Jason Adams" <ROCLIST@ACCESSCONSULTING.CA>; "0sciroccolist"<BR><SCIROCCO-L@SCIROCCO.ORG><BR>Sent: Monday, October 28, 2002 1:42 PM<BR>Subject: Re: OT Car crash Physics...<BR><BR><BR>> At 03:57 PM 10/28/2002, Jason Adams wrote:<BR>> >With all the wealth of informed people on the list this shouldn't be too<BR>> >difficult...<BR>> ><BR>> >If I know the weight of the cars, the skid distance, assume complete<BR>> >inelastic collision. how do I work it out...<BR>><BR>> I'm confused here -- what are you trying to work out?<BR>><BR>> >Vf^2 = Vo^2 + 2a(dS)<BR>> ><BR>> >is there something wrong with my assumption for 'a' ?<BR>><BR>> Well, uK isn't a constant... the value you use is a "typical" value.<BR>Every<BR>> tire is different, as is every road surface. And stopping distances on a<BR>> car aren't that easy to compute -- you're dealing with 4 tires on an<BR>object<BR>> that has suspension and that does not have even weight distribution.<BR>><BR>> "Typical" deceleration for a modern automobile is between 8.5 and 10<BR>> ms^-2. Of course, some are far below, and a few are above.<BR>><BR>> Using the 3 braking distance figures I have for the 16V from magazines<BR>(60,<BR>> 80mph from Road & Track, 70mph from Car & Driver), we can compute the<BR>> average a for the 16v's braking:<BR>><BR>> 60mph 150 feet 7.87g<BR>> 70mph 196 feet 8.19g<BR>> 80mph 257 feet 8.16g<BR>><BR>> The average of those 3 stops is 8.08g. The 8V (158 feet from 60, 271 feet<BR>> from 80), averaged 7.61g.<BR>><BR>> Of course, modern tires and brake pads improve those distances: Two years<BR>> ago I did 10 consecutive stops from 60 in my16V with Potenza RE71 tires<BR>and<BR>> Ferodo pads and used my G-Tech Pro to measure the distance. I discarded 3<BR>> runs where the G-Tech did not provide an accurate reading, which left me<BR>> with 7 good runs. I discarded the best and the worst runs, leaving 5<BR>> total. The average of those 5 runs was 131.8 feet, which is an average of<BR>> 8.95g.<BR>><BR>> Jason<BR>><BR>><BR>><BR>><BR>><BR>><BR>><BR><BR><BR>_______________________________________________<BR>Scirocco-l mailing list<BR>Scirocco-l@scirocco.org<BR>http://neubayern.net/mailman/listinfo/scirocco-l</BLOCKQUOTE><p><br><hr size=1>Do you Yahoo!?<br>
<a href="http://webhosting.yahoo.com/ ">Y! Web Hosting</a> - Let the expert host your web site
--0-1954395481-1035850515=:33059--