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car's trajectory

D

Don Stauffer in Minnesota

Jan 1, 1970
0
Mark-T wrote:

) Thanks, I didn't know that. It wasn't supposed
) to be a trick.
)
) What about front vs. rear wheel drive? It
) might seem to make no difference, but
) imagine the front wheels rotated a full
) 90* to the left... there will be a
) difference in path, depending on front
) or rear drive.

Assuming perfect grip, there will not be a difference.

SaSW, Willem
Good point. Pure geometry can only handle very low speeds where there
is virtually zero slip angle. At any speed where there is non-zero
centrifugal force, all tires must run a slip angle, so tire does not
roll on a line perpendicular to its axle. Ackerman angles are
computed for such virtually zero slip angles.

Well, maybe I shouldn't have said ALL wheels run a slip angle. In
certain racing cars that are heavily oversteering, sometimes they run
with the left front wheel actually up in air, so that wheel is doing
virtually nothing, and the geometry is the same as a weird tricycle.
 
H

hagman

Jan 1, 1970
0
Consider a vehicle of axle-to-axle length L,
and left to right wheel separation W (though
I don't believe this matters).

Assume a X-Y co-ordinate plane, the origin
located at the center of the rear axle.

What is the car's forward trajectory,
if the front wheels rotate left at angle U?
What is its reverse trajectory?

What if the car has front wheel drive?

Why does one park into a space by backing
up, rather than forward?

** BONUS CREDIT **

Use your answer above, to solve the parallel
parking problem.

Mark

With very rough approximations, we simply note that the back axis
wheels remain straight, hence the (infinitely thin) car is always
tangent to the trajectory of the back wheel(s).
Now a circular arc that is tangent to the final parking position of
the back wheel (essentially tangent to the sidewalk) will surely
intersect the car immediately behind you, but can easily pass the car
in front of you. Hence you must have entered the space moving
backwards.

hagman
 
R

Rich Grise

Jan 1, 1970
0
Why does one park into a space by backing
up, rather than forward?

Because it's so much easier to get into the space with only about
3-5 turns, rather than 18 or 20 or so. :)

Cheers!
Rich
 
W

Willem

Jan 1, 1970
0
N8N wrote:
) While it would be technically difficult to rotate the wheels of either
) car 90 degrees in either direction, it'd be *more* technically
) difficult to rotate the wheels of a FWD car that far...

I have seen footage of a car that had four wheel drive, and where the
wheels could rotate 90 degrees. All four of them. This meant the driver
could drive sideways (crab like) out of a parallel park situation.


SaSW, Willem
--
Disclaimer: I am in no way responsible for any of the statements
made in the above text. For all I know I might be
drugged or something..
No I'm not paranoid. You all think I'm paranoid, don't you !
#EOT
 
Consider a vehicle of axle-to-axle length L,
and left to right wheel separation W (though
I don't believe this matters).

Assume a X-Y co-ordinate plane, the origin
located at the center of the rear axle.

What is the car's forward trajectory,
if the front wheels rotate left at angle U?
What is its reverse trajectory?

What if the car has front wheel drive?

Why does one park into a space by backing
up, rather than forward?

** BONUS CREDIT **

Use your answer above, to solve the parallel
parking problem.

Mark


Sorry for my poor knowledge of the English language.

To calculate the turning radius you have got to determine the steer
angle of the outside wheel, SAo. According the Ackermann principle
this can be done on the basis of the steer angle of the inside wheel
(SAi) with the formula :

SAo = arccot ( (tw/wb)+cot SAi)

where tw is the track width of the vehicle and wb the wheelbase

The turning radius odf the outside front wheel (Rof) is then :

Rof = wb/sin SAo

A detailled description of the formulas is given in Daily e.a.,
Fundamentals of Traffic Crash Reconstruction, ISBN 1-884566-63-4

I hope this gives you a perspective for a solution

Hubert Ruypers
 
R

Rich Grise

Jan 1, 1970
0
In other words, the same reason that forklifts steer with their rear
wheels.

Think about it.

I had assumed that forklifts were like that because it'd be much more
expensive to make a steerable axle that also supports 40,000 lbs. ;-)

Cheers!
Rich
 
K

krw

Jan 1, 1970
0
[email protected]>, [email protected]
says...
Thanks, I didn't know that. It wasn't supposed
to be a trick.

What about front vs. rear wheel drive? It
might seem to make no difference, but
imagine the front wheels rotated a full
90* to the left... there will be a
difference in path, depending on front
or rear drive.

There is a difference in path because each wheel is not the same
distance from the center of the turn. If there wasn't one center
point of the turn, either the tires will skid or the car will come
apart. ;-)
 
J

JosephKK

Jan 1, 1970
0
Richard wrote:
) Phil Carmody said:
)> Yes. The car will go nowhere. Assume perfect grip and
)> infinitely narrow incompressible tyres, perhaps?
)
) In which case it will dig itself further and further into the tarmac, until
) the engine dies.

If the tyres are infinitely narrow, wouldn't the car drop like a brick ?


SaSW, Willem

Only until the frame hit the ground.
 
J

JosephKK

Jan 1, 1970
0
[email protected]>, [email protected]
says...

There is a difference in path because each wheel is not the same
distance from the center of the turn. If there wasn't one center
point of the turn, either the tires will skid or the car will come
apart. ;-)

Please make diagrams to support this extraordinary claim.
 
K

krw

Jan 1, 1970
0
Please make diagrams to support this extraordinary claim.
Diagrams in a text-only medium are problematic. ...but think about
it.
 
R

Rich Grise

Jan 1, 1970
0
Please make diagrams to support this extraordinary claim.

I can give you an example - have you ever driven a car with a misaligned
front end? When you turn, the misaligned tire squeals. If the tires had
infinite sliding friction, and the rubber was stronger than steel, then
something would break under those conditioins.

I bought a used '75 Plymouth Gran Fury in the early 1990s, and the front
end was way screwed up, but the car was $500.00. The front end needed an
overhaul, and I was lucky my neighbor at the time was a car guy. The first
thing it needed was an A-arm bushing, then tie rods, pitman arm, and idler
arm, and the tires stopped scuffing (squealing).

Hope This Helps!
Rich
 
R

Rich Grise

Jan 1, 1970
0
Ah right....I usually use speed and the handbrake for that!

Like the one in that commercial? You're lookng at an empty parking
spot, you hear squealing tires, and somebody does a half-donut and
winds up exactly in the spot. The voice-over says, "[car brand] -
for people who can park themselves" - clearly competing with those
newfangled self-parking cars.

The maneuver was pretty impressive, if it can really be done. ;-)

Cheers!
Rich
 
J

JosephKK

Jan 1, 1970
0
Diagrams in a text-only medium are problematic. ...but think about
it.

So post it to abse or on a web page. Guessing from the headers you
are likely to be able to do both. Failing that maybe someone else
here could do either or both, given transfer of the original.
 
J

JosephKK

Jan 1, 1970
0
I can give you an example - have you ever driven a car with a misaligned
front end? When you turn, the misaligned tire squeals. If the tires had
infinite sliding friction, and the rubber was stronger than steel, then
something would break under those conditioins.

I bought a used '75 Plymouth Gran Fury in the early 1990s, and the front
end was way screwed up, but the car was $500.00. The front end needed an
overhaul, and I was lucky my neighbor at the time was a car guy. The first
thing it needed was an A-arm bushing, then tie rods, pitman arm, and idler
arm, and the tires stopped scuffing (squealing).

Hope This Helps!
Rich

Presumably truthful, somewhat interesting, but does not answer my
question. Does the steering configuration truly provide a single
center of rotation?
 
K

krw

Jan 1, 1970
0
So post it to abse or on a web page. Guessing from the headers you
are likely to be able to do both. Failing that maybe someone else
here could do either or both, given transfer of the original.

Nope. No access to abse (any binaries). Don't have a web page.
Think about it some more...

If two points are moving in a circle around different points, is
there not a non-zero velocity between them? That tends to break
solid things containing those two points.
 
K

krw

Jan 1, 1970
0
Rich did answer but in his way and it was basically correct.
:)
let me try....
If the tires do not have a slightly different turning angle,
they will not follow the correct path and will squeal
and start to tear apart stuff the longer you let it go.
(slowly but surely)
anyways...
The outside tire follows a long path,
The inside one a shorter path.
making 2 circles.
The outer circle has a smaller angle of turn
than the inner circle has.
Most cars do make the steering axles do such by changing the camber angle
http://en.wikipedia.org/wiki/Camber_angle and a
http://en.wikipedia.org/wiki/Caster_angle
You can change the angle at which the wheel is turning
compared to equal instead.

Alignment is the key word.
and BTW:
alignment will pretty much never cause a shake while
going straight.
that is of course the balancing.
:)

Nonsense. Just went through that on SWMBO's car.
 
K

krw

Jan 1, 1970
0
The 2 points are not moving around 2 points,
They are both moving around one point.

Exactly the issue at hand. JKK doesn't believe it.
One is moving faster WRT the other.

For some definitions of "faster". ;-)
2 different size circles centered around the same point.
that is all.

Tell it to the physics non-believer.
 
K

krw

Jan 1, 1970
0
It is not nonsense.
It is mechanical law.

NO in fact, it's not. Your theory is lacking reality.
You best check it out a bit more than you did.
:)

Wrong again. Done deal. Alignment fixed everything balancing
didn't.
 
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