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Old 12-01-2007, 08:45 PM
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Steering angle set up.

I am setting up my chassis/body/suspension and need to define some clearances for the front wheels and fenders.
What do think the minimum steering angle should be for the average road car? I am talking about the angle the wheel will make to the centerline of the car at full lock as you look down on the car.

My very rough guess would be about 35 degrees. Does this sound right? Thanks for any help.

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Old 12-01-2007, 08:53 PM
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Coule you go out to your daily driver with a protractor and actually measure a wheel at lock? That's what I would do.

Longer vehicles need more angle to keep within an acceptable turning radius.

Later, mikey
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Old 12-01-2007, 10:13 PM
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What exactly are you ralking about, caster or SAI (KPI)?

Brian
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Old 12-02-2007, 12:36 AM
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Check this thread from another forum out..
http://www.eng-tips.com/viewthread.c...=104278&page=1

It may have some useful math in it.

Later, mikey
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Old 12-02-2007, 07:48 AM
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Mike
thanks for the reply. I did read that thread before I posted but they talk about angles from 32 to 54 deg. Maybe someone here could give me a smaller range or have some figures from a drawing. I tried measuring my truck but it was difficult to get the angle with any accuracy, I could of been 5 degrees out, also turning radius on my truck is huge.

To be honest I was hoping people who have set up their own chassis/suspension systems would know this figure.

Brian - I am not talking about caster OR steering axis inclination, simply the angle the wheels make at full lock. I am not sure how else to describe it, I have only seen it written as steering angle.
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Old 12-02-2007, 09:48 AM
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Ahhhhh, hints the "larger vehicle needing more" I got cha,

Brian
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Old 12-02-2007, 09:50 AM
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Of course, you want to use another car as an example only after you have driven that car and find it acceptable. You wouldn't want to use a PT Cruiser as an example, they take a friggin acre to turna round!

Brian
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Old 12-05-2007, 06:50 AM
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Hey Brian,

NO KIDDING!!! My wife has an '02 PT, and I can NEVER park the thing without making two tries. It's horrible. Gets lousy milage too, and the rear drum brakes make a God awful racket no matter what type of shoes it's got.

On the up side, it is ugly. Voted #1 in "Click and Clack's" "10 ugliest cars made" contest a year ago. The winning nomination went something like, "It's always so dissapointing to see one that the flames are only painted on."

When we bought it, I thought if she wanted a car that looked like a '37 Ford, we should buy, for less money, a finished street rod and just drive the crap out of it. She thought otherwise...

LOL, but, I digress, what was the original question in this thread? Oh yeah, steering angle. That'll be determined by the spindles steering arm length, pitman arm and the box or rack. Unless one is fabricating all those, it's kinda hard to say before hand, isn't it?

Brian
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Old 12-05-2007, 07:33 AM
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What type of car are you building? I think This would help get more precise info. Also straight axle or mustang II,ect.
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Old 12-05-2007, 08:27 AM
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Alright... looking through your past posts, I see you asking about '32 Ford body weight, so I'll make the slightly less than wild guess that you're looking at building a '32 Ford, and I'll use it as an example for even if you're not.

The '32 Ford has a 106" wheelbase. I happen to like my for parking, and it has a minimum turn radius of 17'. (204") Let's figure out what the turn angle of the inside wheel is using a bicycle mode for simplicity.

Here's how it works: The bicycle model allows us to neglect considerations of tire scrub and track width, The most inboard point on your car is the inside rear wheel, so we'll assume that's the point of measuring the turn radius, and since you're interested in a maximum turn angle for your wheel and the inside front wheel turns more than the outer wheel of a turn, it works out that you're only interested in both inside wheels anyway.

The center of your turn is defined by the intersection of the wheel axis of your front and rear wheels. That means you have a simple right triangle. The angle you are looking for would be found by:

angle = 90 - arctan(turn radius/wheelbase) = 90 - arctan (204/106) = 90-63
27 degrees
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Old 12-05-2007, 06:14 PM
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The PT Cruiser, of course, was designed after the 37 Chrysler, not the 37 Ford!
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Old 12-05-2007, 07:09 PM
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Slipangle

great info thanks for the post. So what your saying is that you think a good size turning radius is 17 feet and then you just worked backward calculating for that turn circle and came up with a steering angle of 27 degrees.

So in my case with a wheelbase of 140"(no I am not building a 32 Ford!) it would be

90-arctan(204/140)=34.5 deg. approx. That's great, I sort of planned for 35 deg.
Thanks again for the help.

Axle will probably be a Heidt's superide or similar, would it be OK for me to assume that this kind of front suspension will accomodate this angle?
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Old 12-05-2007, 09:55 PM
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The minimum turning radius is determined by the travel of the outside front tire. With a desired minimum turning radius of 17 feet, for a wheelbase of 140 inches, the necessary steer angle would be:

angle = arcsine (140/(12 x 17)) = 43.3 degrees.

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Old 12-06-2007, 04:27 AM
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Billy

thanks for the info. Here is how I understand your reply as opposed to slipangle's reply.

You are talking about a curb to curb turning distance i.e. the total circle scribed by the car (minus body overhangs) when turning a full circle at full steering lock. Which would be 34 feet.

The other post was the calculation of the inside wheel and the turning circle IT scribes when turning a circle at full steering lock. So to calculate the actual curb to curb distance when using the other formula I would have to add twice the TRACK of my vehicle to the 34 feet?
Does this sound correct to you?

If so I guess my next question is this -
What is a suitable turning radius for a vehicle with a wheelbase of 140" ? (I expect it to be on the large side but not unwieldy)
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Old 12-06-2007, 07:26 AM
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Quote:
Originally Posted by scrimshaw

You are talking about a curb to curb turning distance i.e. the total circle scribed by the car (minus body overhangs) when turning a full circle at full steering lock. Which would be 34 feet.
That's a diameter of 34 feet and a turning radius of 17 feet. This is the figure quoted in Motor Trend's road tests. I would suggest that you take a look at some back issues and thereby get an idea of what's reasonable. Front wheel drive cars generally have a larger turning radius than rear wheel drive cars, so keep that in mind as you look at the road tests.

Incidentally, I was assuming true Ackermann steering, which would mean that the front inside wheel, assuming a front track of 60 inches, would turn through 56.2 degrees. With parallel steering (no Ackermann), both wheels would turn 50.1 degrees.

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