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How to prove a car must turn in a curve? 
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#20
Jan2113, 02:59 PM

P: 163

Sorry about the confusion: thank you so much for you guys help, I did not expected this much responds from such badly expressed question: for the math: http://tinypic.com/view.php?pic=ychgp&s=6 (I'm aware of the car differential btw) for the car physics: my goal here is to try to understand why cars turn the way they do, or why they are designed the way they are for turning, and I would like to know if the following is true: each velocity are "designed" to move at a certain direction, velocity, such that they "resemble" a rotating rigid body. where direction of wheels are normal to their instantaneous center of velocity since the back wheels are driving wheel, Im imaginging that when a car turns, it is DUE to the difference in speed between the left and right rear wheels, and we simply orient the front wheels at an angle so that their planes are normal to that center of rotation. in other word , we have 2 restriction of motion : 1. rigid body (if we hold front wheels the same angle relative to the back at all time) 2. no slipping, the wheel's plane are oriented in the same direction of its motion (assume rigid wheels) hence, we are simply "matching" these criteria together, since we know the front wheels position will rotate according to the difference in left and right rear wheels, we make their planes's normal intersect the center of rotation so that it will not slip when that rotation occur. 


#21
Jan2113, 04:03 PM

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You might also remember 1. that half the cars in the world happen to have front wheel drive and 2. that they still go round corners when there is no motive force. I have no idea what this means, i'm afraid. Also, not allowing the wheels to slip in your model is missing the major part of the way the car steers. You may as well propose that the steering is achieved by curved rails  which is relatively easy to analyse. 


#22
Jan2113, 04:18 PM

P: 2,048

The reason why you are having such problems is that you are eliminating the mechanism by which cars turn at anything above a crawling pace. Cars turn by slipping, by assuming no slip you have created a situation that is irreverent to real world vehicles. Also only consider a front wheel drive case, it's simpler as you can ignore the rear wheels as they are free to spin at whatever speed they need to. 


#23
Jan2113, 05:27 PM

P: 163

Ok....I guess I am very confused the... so how should I go about understanding the rotation of the car? what is the cause and what is the effect here? I am very confused... 


#24
Jan2113, 05:28 PM

P: 163

what do you mean by "crawling"? I'm confused as in, if wheels can slip, then why can't the front wheels just move at the same angle so that backwheels "slip" and the car turn translationally without rotating? if not, what are the restriction on this slipping of the wheels? thanks! 


#25
Jan2213, 04:28 AM

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I'm sorry but some of your choices of words make it very difficult to understand what you mean (I also found your diagram a bit confusing, I'm afraid.) I just looked in an old book of mine on sports cars by Colin Chapman  designer of early Lotus cars. It reminded me of the term 'Ackerman Steering' Google this term and you will find loads of stuff to help you. 


#26
Jan2213, 05:18 AM

P: 2,048

http://www.youtube.com/watch?v=d01VBR5nKPw 0:070:10 At very very low speeds, the car turns in the direction that the wheels are pointed. So your turn centre is in line with the rear wheels. Ie, draw a line perpendicular through the rear wheels and perpendicular to the front wheels. Unless you have a geometry that allows the inner and outer wheels to take different paths (because the inner wheel is taking a shorter radius) you will scrub the tyres. Reference: Ackermann steering angle At higher speeds the tyre tread distorts meaning the tyre is travelling in a different direction than the contract patch. The angle between these two is the slip angle. This slip angle induces a cornering force which acts to turn the car into the turn. This is the crucial thing. The lateral force caused by the turn induces a slip angle in the rear wheels. So the rear wheels are actually contributing to the turn. The centre of the turn (instantaneous centre) is now perpendicular to the slip angles front and rear. When the front slip angle exeeds the rear, the car will understeer. When the rear slip angle exeeds the front, the car will oversteer. In your scenario, the car will permanently and terminally understeer as you have no slip angle on the rear. I've bolded several words/concepts above. You should try to read up on them before approaching this again. If you have any questions on them, just ask. 


#27
Jan2213, 05:53 AM

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P: 7,126

wiki_ackermann_steering Even in the idealized case of no tire slipping and perfect ackermann steering, you still have some of the turning force provided by the rear tires, due to the centripetal acceleration of the car resulting in a reactive outwards force on all 4 tires, with an equal but opposing inwards (centripetal) force from the tires (a Newton third law pair of forces, you also have outwards force of tires on pavement, and equal and opposing inwards force of pavement on tires). The ratio of force between front and rear depend on weight distribution of the car (assuming constant speed). Image of an idealized example of a car turning, no tire slipping, and perfect ackerman steering setup: 


#28
Jan2213, 02:03 PM

P: 3,143

I have another drawing that shows how all four wheels can be tangents to the curve they are following if the front two wheels are at different angles...



#29
Jan2213, 03:34 PM

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PS No one has yet mentioned the technique of additional rear (four) wheel steering which has been used for high end cars. 


#30
Jan2213, 06:10 PM

P: 163

And are these "inward" forces reaction forces normal to the planes of the front wheels caused by the "forward" forces on front wheels by the back wheels? 


#31
Jan2313, 02:50 AM

P: 3,143

The cars inertia will try to make it go in a straight line (a tangent to the curve). The front wheels point in a different direction so they provide a sideways force on the front of the car.
In effect inertia is trying to drag the froont wheels sideways and it's friction that provides the sideways force making the car turn. On ice you go straight on! The same applies to an un powered soap box cart so makes no difference if the car is front or rear wheel drive. 


#32
Jan2313, 04:49 AM

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#33
Jan2313, 05:18 AM

P: 163

No I haven't yet, and I am going to just discard al preassumptions I have at this moment and read them....guess that's my best bet. I thought I knew, but I don't, not one bit, my arrogance has cost me 2 days of mental torment. And it is time I liberate myself , I will not ask anymore questions until I have squeeze every last bit of all the resources made available to me by you all good folks. 


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