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Given width, wheelbase, length and turning circle , is there a mathematical formula that determines the minimum size space you can parallel park in? Assume cars are rectangular with no sticking out bits.
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Given width, wheelbase, length and turning circle ,
What about track? Or do you assume track =width?
Clearly you've never clapped eyes on a Nash Metroplitan ;-)
Hawkeye
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Stranger in a strange land
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Also the width of the cars at either end of the space (or at least the one you start next to) would surely effect how long a space you needed to be able to get right against the kerb.
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you forgot to mention the hand brake
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The limiting factor for having an infinite number of "to and fro" goes is probably the diagonal length of the car, assuming the three cars involved are the same width. So, using pythagoras.
gap = square root of (length squared plus width squared).
In practical terms you aught to add a very small ammount to the figure produced, so you don't jam with no room for manouvre.
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Surely theoretically you can get into any space as long as there is even a fraction of an inch clearance? What really matters is how many backwards and forwards goes you are prepared to use to make the exercise worthwhile.
I once extricated my Volvo 240 from a space that had only about 4" of clearance. I was boxed in with no chance of contacting the culprit, so it was worth about 50 movements for the sake of being able to get home that evening.
Also I wasn't too careful about avoiding bashing his car with the tow bar.
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If the track is equal to the width of the parking car, and the width of the parked cars is also equal to the track
E = wheelbase
t = track
alpha = maximum steering angle
b=sqrt( (E/sin(alpha))^2 - ((E/tan(alpha))-t)^2)
If I put more thought into it, the expression for b could probably be simplified a bit - but it is the weekend!!
The distance between the two parked cars has to be
Front overhang + b + rear overhang + some skill dependent clearance!!
This is actually a slight over-estimate because the front overhang is at an angle as it swings past the corner of the parked car - but it is probably a small error.
Number_Cruncher
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mmm can't get head round that tonight.
Supposing you're parking a handcart with only two wheels located centrally - doesn't that make E = zero ?
Also can you write equations using MS Office or do you need a separate package?
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Supposing you're parking a handcart with only two wheels located centrally - doesn't that make E = zero ?
The equation isn't any use for handcarts! It assumes that your steering is done conventionally, rather than by pivoting the entire axle.
It's much easier to understand the equation with a drawing of the geometry, which, alas, I can't include in a post!
You can play about with this type of equation in Excel easily enough, although for choice, I would either use MATLAB, or code it up in Fortran. It depends what you want to do with the equation really - Excel is good for seeing what each term is doing, but is poor if you want to use complex numbers, or useful things like Fourier Transforms. MATLAB is very powerful, has great plotting and graphics, but is not cheap (although the Student version is good value), and Fortran can be had cheaply, and can produce very fast programs, but it takes more programming skill to use.
If you prefer to work with symbols rather than numbers, the MAPLE or Mathematica may be more suitable. Sometimes I do cheat, and I paste Mathematica output into my MATLAB code!
Number_Cruncher
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N
FORTRAN! Ooh I've come over all nostalgic. If it begins with I,J,K,L,M or N it must be an integer!
JH
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>>I've come over all nostalgic
I don't use Fortran anywhere near as much as I used to. The newest releases of MATLAB now can run very fast code - and of course you can just cut snippets of code and run them at the command prompt instead of compiling and linking, which makes de-bugging much simpler. You need to be doing something a bit special now to make the extra effort of Fortran coding worthwhile.
Number_Cruncher
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>>Fortran can be had cheaply
Indeed. This compiler is free
www.silverfrost.com/32/ftn95/ftn95_personal_editio...p
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Think you need to include a term for the camber of the road and differential vehicle heights. Outside my house is a sleeping policeman and using rhe slope of this to lift the front end by 6 inches means that my bumper will pass over the next cars bumper thus reducing the required space. Need to brake sharply as the car runs down the slope of the sleeping policeman though or face embarassing chats with the neighbours.
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Probably needs some practical testing:
www.youtube.com/watch?v=fV35mVrwhUQ
(Where's Russ?)
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(Where's Russ?)
Russ Swift was performing in a couple of those clips.
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">Russ Swift was performing in a couple of those clips.<"
Yes Dave .....
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Rectangular cars are more difficult to park than cars with rounded corners. If they are really rectangular - Volvo 240? - they certainly need a proportionately longer space to be able to park. Indeed a car of the same overall length and width which is rectangular won't let itself be driven into a space that one with rounded off corners will fit into.
We all know this from experience. In practice though a car can be shoehorned into a remarkably small space given patience, practice and a measure of ruthlessness. Nothing to do with mathematical formulae of course, just intuitive geometry and nice bouncy modern bumpers... Don't do it next to DD's car though or he'll come after you with a baseball bat.
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Don'tdo it next to DD's car though or he'll come after you with a baseball bat.
Size 11 steel toe capped boots, IIRC ;o)
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Size 11 steel toe capped boots, IIRC ;o)
Yes, I was right:- www.honestjohn.co.uk/forum/post/index.htm?v=e&t=41...0
:o) :o)
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What do you mean, :o) ;o)
:>{ more like, or even :~0
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I think there are two underlying threads here.
1) is the smallest space you can get the car into in one go. Obviously how rounded the corners are is crucial here.
2) is whether there is in fact a lower limit if you are willing to make hundreds of tiny movements, gradually inching the car in sideways, gaining a tenth of an inch each time.
You are in a tight space. Go back until just touching the car behind. Full lock (never mind about scrubbing the tyres). Go forward half an inch. Full lock the other way. Go forward half an inch. Full lock, go back half an inch, etc etc etc etc.......
You have to keep the car basically parallel all the time, and not try to be too clever by gaining too much on each go. You need to "waste" half of every movement in order to straighten up ready for the next move.
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Is anyone else picturing a scene from an Austin Powers movie? ;-)
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or the Austin Maestro (or Montego?) ad?
JH
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