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OT Car crash Physics...
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I would estimate the collision occured at 30 degrees to the jag's =
midline i.e. if motorcycle was headed due north, jag was pointing 150 =
degrees.
I expect the front was rotated 2.4 m and car pivioted on driver rear =
wheel with a 2.6m wheelbase thats ~43 degrees.
The force of impact bent the 'frame' of the jag, i.e. driver's fender =
(opposite to side of impact) was also crinkeled.
I expect brakes were released at instant of impact.
I found values of u to be .75 for static and .72 for kinetic.
if Ff =3D u Fn then:
Ff =3D .72 * 9.81m/s^2 * 2480kg =3D 17517 Newtons
and
1/2 mVo^2 =3D r * F * Sin(theta) Kinetic energy =3D torque applied
then
Vo =3D [(2 * r * F * Sin(theta))/m]^ -2
where r is wheelbase, m is mass of motorcycle&driver, theta is 43 =
degrees, and F is force as calc above
i solve Vo to be 16 m/s or 58km/h=20
this seems way too slow to me.... what am I missing???
Jason Adams
Fahrvergnugen Forever! ;)
84 rocco 16v
93 320i
98 Z71
----- Original Message -----=20
From: Brian Haygood=20
To: Jason Adams ; 0sciroccolist ; Jason ; Scirocco-Al=20
Sent: Monday, October 28, 2002 4:15 PM
Subject: Re: OT Car crash Physics...
I used to do accident reconstruction, so maybe I can help.=20
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. =20
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. =20
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.=20
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. =20
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.=20
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). =20
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. =20
Jason Adams <roclist@accessconsulting.ca> wrote:=20
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"=20
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 =3D 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
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<DIV><FONT face=3DArial size=3D2>I would estimate the collision occured =
at 30=20
degrees to the jag's midline i.e. if motorcycle was headed due north, =
jag was=20
pointing 150 degrees.</FONT></DIV>
<DIV><FONT face=3DArial size=3D2></FONT> </DIV>
<DIV><FONT face=3DArial size=3D2>I expect the front was rotated 2.4 m =
and car=20
pivioted on driver rear wheel with a 2.6m wheelbase thats ~43=20
degrees.</FONT></DIV>
<DIV><FONT face=3DArial size=3D2></FONT> </DIV>
<DIV><FONT face=3DArial size=3D2>The force of impact bent the 'frame' of =
the jag,=20
i.e. driver's fender (opposite to side of impact) was also=20
crinkeled.</FONT></DIV>
<DIV><FONT face=3DArial size=3D2></FONT> </DIV>
<DIV><FONT face=3DArial size=3D2>I expect brakes were released at =
instant of=20
impact.</FONT></DIV>
<DIV><FONT face=3DArial size=3D2></FONT> </DIV>
<DIV><FONT face=3DArial size=3D2>I found values of u to be .75 for =
static and .72=20
for kinetic.</FONT></DIV>
<DIV><FONT face=3DArial size=3D2></FONT> </DIV>
<DIV><FONT face=3DArial size=3D2>if Ff =3D u Fn then:</FONT></DIV>
<DIV><FONT face=3DArial size=3D2></FONT> </DIV>
<DIV><FONT face=3DArial size=3D2> Ff =3D .72 * 9.81m/s^2 * 2480kg =
=3D 17517=20
Newtons</FONT></DIV>
<DIV><FONT face=3DArial size=3D2></FONT> </DIV>
<DIV><FONT face=3DArial size=3D2>and</FONT></DIV>
<DIV><FONT face=3DArial size=3D2></FONT> </DIV>
<DIV><FONT face=3DArial size=3D2>1/2 mVo^2 =3D r * F * =
Sin(theta) =20
Kinetic energy =3D torque applied</FONT></DIV>
<DIV><FONT face=3DArial size=3D2></FONT> </DIV>
<DIV><FONT face=3DArial size=3D2>then</FONT></DIV>
<DIV><FONT face=3DArial size=3D2></FONT> </DIV>
<DIV><FONT face=3DArial size=3D2>Vo =3D [(2 * r * F * Sin(theta))/m]^ =
-2</FONT></DIV>
<DIV><FONT face=3DArial size=3D2></FONT> </DIV>
<DIV><FONT face=3DArial size=3D2>where r is wheelbase, m is mass of=20
motorcycle&driver, theta is 43 degrees, and F is force as calc=20
above</FONT></DIV>
<DIV><FONT face=3DArial size=3D2></FONT> </DIV>
<DIV><FONT face=3DArial size=3D2>i solve Vo to be 16 m/s or 58km/h=20
</FONT></DIV>
<DIV><FONT face=3DArial size=3D2></FONT> </DIV>
<DIV><FONT face=3DArial size=3D2>this seems way too slow to me.... =
what am I=20
missing???</FONT></DIV>
<DIV><FONT face=3DArial size=3D2></FONT> </DIV>
<DIV><FONT face=3DArial size=3D2></FONT> </DIV>
<DIV><FONT face=3DArial size=3D2></FONT> </DIV>
<DIV>Jason Adams<BR>Fahrvergnugen Forever! ;)<BR>84 rocco 16v<BR>93 =
320i<BR>98=20
Z71<BR></DIV>
<BLOCKQUOTE=20
style=3D"PADDING-RIGHT: 0px; PADDING-LEFT: 5px; MARGIN-LEFT: 5px; =
BORDER-LEFT: #000000 2px solid; MARGIN-RIGHT: 0px">
<DIV style=3D"FONT: 10pt arial">----- Original Message ----- </DIV>
<DIV=20
style=3D"BACKGROUND: #e4e4e4; FONT: 10pt arial; font-color: =
black"><B>From:</B>=20
<A title=3Dred8716v@yahoo.com href=3D"mailto:red8716v@yahoo.com";>Brian =
Haygood</A>=20
</DIV>
<DIV style=3D"FONT: 10pt arial"><B>To:</B> <A =
title=3Droclist@accessconsulting.ca=20
href=3D"mailto:roclist@accessconsulting.ca";>Jason Adams</A> ; <A=20
title=3Dscirocco-l@scirocco.org=20
href=3D"mailto:scirocco-l@scirocco.org";>0sciroccolist</A> ; <A=20
title=3Djason@scirocco.org =
href=3D"mailto:jason@scirocco.org";>Jason</A> ; <A=20
title=3Dscirocco-Al@insight.rr.com=20
href=3D"mailto:scirocco-Al@insight.rr.com";>Scirocco-Al</A> </DIV>
<DIV style=3D"FONT: 10pt arial"><B>Sent:</B> Monday, October 28, 2002 =
4:15=20
PM</DIV>
<DIV style=3D"FONT: 10pt arial"><B>Subject:</B> Re: OT Car crash=20
Physics...</DIV>
<DIV><BR></DIV>
<P>I used to do accident reconstruction, so maybe I can help.=20
<P>What do you mean by "back remaind stationary" If the car was =
hit head=20
on and collapsed 8 feet with the back wheels not moving, you are then =
left=20
with a jaguar half the original size. Since there is nothing to =
keep the=20
back wheels in their place, this wouldn't happen anyway. The jag =
would=20
move a lot before/while collapsing that way. =20
<P>Did the noble jag pivot about it's rear wheels when struck? =
Maybe so,=20
if it was hit at more of an angle than your description lead me to=20
believe. The problem is that the rear of the vehicle would never =
have=20
stayed in place - at least one rear wheel has to move for the car to=20
pivot. How much? At what angle? All of this =
stuff is=20
figured out by measuring skid marks on the pavement, their exact =
direction=20
(for each wheel) and mapping the movement of the car onto these =
marks. =20
<P>Let me just tell you how we would work an accident reconstruction =
like=20
this. We would measure the degree to which the jaguar is crushed =
in and=20
compare that with crash test data on the jag to determine how much =
energy it=20
takes to crush that car that much. From that you can directly =
determine=20
the difference in the speed of the two objects with pretty good=20
accuracy. Since you know the jag was stopped, that's all you =
need.=20
<P>Another approach would be to measure the displacement of the jag =
and=20
motorcycle after the crash, and see how much energy they used up in =
coming to=20
rest. That's more the approach you seem to be going for. =
There are=20
several catches.=20
<P>Easiest way is to assume that the car and bike became one after =
they=20
hit. If that combined mass came to rest after travelling 2.4m, =
then all=20
you need is some coef of friction and you are set. =
The=20
problem is again in the crushing of the car. The distance the =
combined=20
car/bike unit moves has to be taken as the distance the center of mass =
of the=20
unit moves. If the jag didn't crush much, then it doesn't =
matter, but if=20
it did, then the center of mass is moving a lot less than the front =
wheels=20
are. =20
<P>So if the jag didn't crush much, you can still procede, but what =
angle was=20
the impact? If the brakes of the jag were fully applied =
throughout the=20
collision such that the wheels were always locked up, you don't really =
care=20
much (as long as you have corerectly figured out the displacement), =
but no one=20
I know holds their brake pedal down to the floorboard when waiting at =
a light.=20
<P>If you have gotten through all of this, then you only have to =
figure out=20
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=20
(imagine that). =20
<P>So here's your estimate: if a jag is moved 2.4m by =
something a=20
tenth its size, then the bike was absolutely hauling ass. Hope =
that=20
helps. =20
<P> <B><I>Jason Adams <roclist@accessconsulting.ca></I></B> =
wrote:=20
<BLOCKQUOTE=20
style=3D"PADDING-LEFT: 5px; MARGIN-LEFT: 5px; BORDER-LEFT: #1010ff 2px =
solid">M1=20
is Jaguar XJS curb weight 4808lbs driver 165lbs ---> 2260kg<BR>M2 =
is=20
Yamaha Motorcycle 360lbs + 125lbs ----->220kg<BR><BR>Jaguar is =
stationary=20
(waiting to turn left) and is struck by motorcycle<BR>headlong in =
the pass=20
front fender.<BR><BR>Front of car is displaced 8 feet (2.44m) back =
remains=20
stationary, wheelbase<BR>is 102" (2.59m)<BR><BR>uK is a coefficient =
I found=20
on the net.<BR><BR>I want to calculate the inital velocity of the=20
motorcycle.<BR><BR>and no the motorcycle driver isn't doing so=20
well....<BR><BR><BR><BR>Jason Adams<BR>Fahrvergnugen Forever! =
;)<BR>84 rocco=20
16v<BR>93 320i<BR>98 Z71<BR><BR><BR><BR>----- Original Message=20
-----<BR>From: "Jason" <JASON@SCIROCCO.ORG><BR>To: "Jason Adams"=20
<ROCLIST@ACCESSCONSULTING.CA>;=20
"0sciroccolist"<BR><SCIROCCO-L@SCIROCCO.ORG><BR>Sent: Monday, =
October 28,=20
2002 1:42 PM<BR>Subject: Re: OT Car crash Physics...<BR><BR><BR>> =
At=20
03:57 PM 10/28/2002, Jason Adams wrote:<BR>> >With all the =
wealth of=20
informed people on the list this shouldn't be too<BR>>=20
>difficult...<BR>> ><BR>> >If I know the weight of =
the cars,=20
the skid distance, assume complete<BR>> >inelastic collision. =
how do I=20
work it out...<BR>><BR>> I'm confused here -- what are you =
trying to=20
work out?<BR>><BR>> >Vf^2 =3D Vo^2 + 2a(dS)<BR>> =
><BR>>=20
>is there something wrong with my assumption for 'a' =
?<BR>><BR>>=20
Well, uK isn't a constant... the value you use is a "typical"=20
value.<BR>Every<BR>> tire is different, as is every road surface. =
And=20
stopping distances on a<BR>> car aren't that easy to compute -- =
you're=20
dealing with 4 tires on an<BR>object<BR>> that has suspension and =
that=20
does not have even weight distribution.<BR>><BR>> "Typical"=20
deceleration for a modern automobile is between 8.5 and 10<BR>> =
ms^-2. Of=20
course, some are far below, and a few are above.<BR>><BR>> =
Using the 3=20
braking distance figures I have for the 16V from =
magazines<BR>(60,<BR>>=20
80mph from Road & Track, 70mph from Car & Driver), we can =
compute=20
the<BR>> average a for the 16v's braking:<BR>><BR>> 60mph =
150 feet=20
7.87g<BR>> 70mph 196 feet 8.19g<BR>> 80mph 257 feet=20
8.16g<BR>><BR>> The average of those 3 stops is 8.08g. The 8V =
(158=20
feet from 60, 271 feet<BR>> from 80), averaged =
7.61g.<BR>><BR>> Of=20
course, modern tires and brake pads improve those distances: Two=20
years<BR>> ago I did 10 consecutive stops from 60 in my16V with =
Potenza=20
RE71 tires<BR>and<BR>> Ferodo pads and used my G-Tech Pro to =
measure the=20
distance. I discarded 3<BR>> runs where the G-Tech did not =
provide an=20
accurate reading, which left me<BR>> with 7 good runs. I =
discarded the=20
best and the worst runs, leaving 5<BR>> total. The average of =
those 5=20
runs was 131.8 feet, which is an average of<BR>> =
8.95g.<BR>><BR>>=20
=
Jason<BR>><BR>><BR>><BR>><BR>><BR>><BR>><BR><BR><BR>=
_______________________________________________<BR>Scirocco-l=20
mailing=20
=
list<BR>Scirocco-l@scirocco.org<BR>http://neubayern.net/mailman/listinfo/=
scirocco-l</BLOCKQUOTE>
<P><BR>
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