I would like to see at least 8" rim width up front and at least 9" rim width in the rear.

What do you think??

 

No.

Imagine the cost of replacing tires.

 

Yes, maybe. But look at how good it will look. Aren't tires available individually? I mean, usually we but 4 at a time, but 2 at a time should not be that financially prohibitive. Oh well.

 

On a lightweight car, it would be overkill, worse for foul weather, and add unsprung and rotational weight.

No, thank you.

---jps

 

sputnik, i agree with you that it seems like overkill and would add too much weight but why would it be worse in foul weather? because of the twwo different widths or just wider alltogether?

 

Less weight over larger contact patch = greater chance for slippage on snow.

 

Given the same inflation pressure, wider tires don't give a bigger contact patch, just wider.

 

i'm glad you're here today rich. so why would the wider tires worse in bad weather?

 

Can you please explain this. I would think that a wider tire would have a greater contact area (surface). But who am I.

Thanks

 

I'm sure there's a link out there that explains it better than I can, but a quick search didn't turn up a good one. So, I'll give it my best shot.

Here's what I wrote in another thread:

Many people think that bigger tires have a bigger contact patch. This is not true. The size of the contact patch is related to the pressure in the tire, the weight that is supported, and the strength of the sidewall. Let’s ignore the sidewall effect for a moment. The pressure in the tire is measured in pounds per square inch (or the metric equivalent, if you live in a civilized country). What does that pounds per square inch measurement really mean? If each tire is supporting ¼ of the 3000 lb. weight of the car, that’s 750 lbs. per tire. If the pressure in the tire is 30 lbs. per square inch, that’s 25 square inches of contact space. This does not change no matter what the size of the tire! If the tire is 6 inches wide, the contact patch will be a hair over 4 inches long. If you stick 12 inch wide tires on, the patch will be about 2 inches long (We’ll get more into this later). Stiff sidewalls can hold up a bit of the weight, so the contact patch will be less than is calculated here, but that is clearly a secondary effect. If that didn’t make sense, try this example. Picture an uninflated balloon. If you set it flat on a table, it will have a fairly large amount of surface area touching the table, because there’s very little pressure in the balloon. Now if you blow it up so that it’s really really full (lots of pressure), only a very small part of the balloon will be in contact with the table. Now, press down with your hand on the balloon. More of the balloon touches the table. You could use this balloon as a scale, if you wanted! All you’d need is the size of the part of the balloon that’s in contact with the table and the pressure in the balloon. Neat!

Does that make sense? If not (and I haven't bored you too much), I can try again.

 

Ahhh, Grasshopper is missing one very important variable: volume. His balloon is full of hot air.

Let's take the baloon example again!

Take two different sized ballons, and fill them to the same pressure. Balance the same book on each, and see how big the "patch" on the bottom is. You will find that the bigger balloon has the bigger patch.

The basic thing is that with the same weight and the same tire pressure, you will basically compress the air in the tire X percent. There are many minor variables that also affect this, but that's the basic formula. So, to make an easy example to show how the numbers work out (this is in no way scientific, just a basic example), let's take two tires. Each has the same tire pressure. One has a volume of 40 cubic inches, and the other has a volume of 60 cubic inches. Let's say that with our pressure, the compression will be 25%. That means that with the weight on those two tires, the smaller tire will be compressed to 30 cubic inches, a change of 10 cubic inches, while the bigger tire will be compressed to 45 cubic inches, a change of 15 cubic inches. A bigger change in air volume means a bigger contact patch.

Those of us who design subwoofer enclosures know that the bigger the enclosure, the more "give" that the supporting air has. So, when we adjust the enclosure size, we adjust how much support the sub gets from the enclosure.

---jps

 

The two different widths would hurt handling performance in dry and wet, because the car is a 50/50 balance.

As far as foul weather, wider tires will make you hydroplane quicker, and won't have the same p.s.i. in the contact patch itself (it will be bigger), and so it doesn't do as good of a job in forcing the tire down to the pavement, and the water/snow out. There is a point of diminishing returns, mind you, but generally, putting the same tire, but wider, hurts a little in heavy water/snow.

---jps

 

No need to get insulting, if you're claiming that I'm just blowing "hot air". If I'm wrong I'll admit it. If your intent with that statement wasn't to simply insult, I apologize. If it was, get a friggin' life guy.


Red Herring. The theory is correct, but in practice it's not a large enough effect to make any difference. The tires simply don't compress enough for this to become a factor. The numbers you used are so exagerated in relation to automobile tires, that they are very misleading. Long before you have to worry about this, the sidewall stiffness comes into play, and you've got that to worry about. Your point is at best a third order effect, if that.

Here are some links, :
http://www.howstuffworks.com/question506.htm
http://www.ibmwr.org/otech/tirestuff.html :
" Experimental data indicates that the net pressure distribution at any point in the contact patch depends primarily upon tire pressure. "
http://www.insideracingtechnology.co...sics1,htm.html
"A tire inflated to 30 pounds per square inch (psi) loaded with 300 pounds (lb) flattens to a contact patch of 10 square inches (in2)."


Yep, I've done the same thing. Well, actually bass guitar speakers. This is correct, but again, while it's a primary consideration in this realm, it just doesn't come into play in tires, except as an exceptionally minor factor.

Here are some more links, and I'm done with this thread. It's been argued to death so many places, that I'm sick of the argument. These threads ought to offer a few hours of reading with all of the links, and there's enough information for anyone to come to their own conclusion. No need to re-hash it here.

http://www.miataforum.com/cgi-bin/ul...c;f=3;t=004557
http://www.miataforum.com/ubb/ultima...3;t=005169;p=1

 

Omg, this is such an inane question I can't gather the energy to write out a decent reponse.

225s are plenty fine thank you. Given that this is a sports car, I'd run though tires at a good rate. Paying nearly $200 per tire is not going to make me happy.

Wet weather, you need to have enough pressure to cut though the water to grip the road. Pressure=force/area. Given the same amount of force (weight) and a greater area (tire), you have less pressure on the road. If you can't cut though the water, you hydroplane. If you can't bite hard enough into snow/mud, you spin.

 

Thanks very hevy stuff, but it din't bore me. Thanks man!

 

Ok, I realized this morning that my last response sucked. That'll teach me to post after getting back from the bar (only 1 drink!) and after midnight...

Anyway, here are the 3 things I thought of upon waking this morning.

1. Let's assume you're right, Sputnik. The tire with larger volume (wider, in this case) has a larger contact patch. What happens when you plus size? When you plus size, you dramatically reduce the volume in the tire, so each time you plus size, the contact patch gets smaller! Do you believe that this happens?

2. Back to your version of the balloon example. Why is the contact patch in the smaller balloon smaller? Because the air is compressed, yeilding a higher internal pressure. The contact patch times the pressure still yeilds the weight of the supported item for both balloons. So, since I said that we assume the pressure in each tire is equal, your own version of the balloon example agrees with what I said, we just diasgree on the meaning with regards to the tires. My stated assumption was that the pressure in the tires must be equal. So if you are correct that the volume increase results in less compression of the air, which leads to a lower relative pressure, then we're talking about different things. I would say to increase the pressure in the wider tire to match that of the smaller tire and the contact patches would be the same.

3. Last, I made an unsupported statement that the compression of the air in the tires would be minimal ("The tires simply don't compress enough for this to become a factor. The numbers you used are so exagerated in relation to automobile tires, that they are very misleading."). I can test this in a couple of weeks, even though my comments in #2 make it a moot point. I'm going to be mounting my snow tires, and I can check the air pressure in them before and after I mount them. If your belief that the compression of the air in the tires is a significant factor, the pressure should be a few PSI higher when mounted than when sitting on the ground.


One last thing, I don't want anyone out there to read this and think about lowering the pressure in their tires to get a larger contact patch, and therefore increase grip. That's not how it works! There are a whole host of other issues that effect tire pressure choices, including heat, rollover of the sidewall at the limit, efficiency, and ride quality. If you drop the pressure in your tires, you'll actually deform the tire and get less grip than if you had them properly inflated. This point that I've been discussing is limited to simply the size of the contact patch relative to wheel width.

As for wider tires being worse in bad weather, I'm not as sure on that, so I won't offer an opinion. The explination (for the rain) that seems to make the most sense to me is as follows (link):
"With wide front tire fitments the volume of water confronted is increased and the distance required to drain water laterally across the tread pattern is greater. The potential to aquaplane is consequently higher."

 

The contact patch doesn't change area wise with wider tires, it only changes shape. A larger rim width doesn't necessairly mean you have to go with a wider tire. 205/50/16 tires for example have a range or rim widths that are acceptable to use. Wider rims would reduce sidewall flex and improve lateral handling if used with same size tire no?

 

That's how I understand it, Quick_lude. However, you're getting into an area that I don't claim to understand fully.

 

That depends on what you mean by "plus size", as in it could mean doing a couple of things. My main point was between two tires with the same wheel diameter, same final circumference. Only the widths would change, henceforth the volume will increase.

As for the "numbers" in my example, they were not accurate. I just put numbers to the "formula" to show how things would change, and I exagerated so that it was clear. That's what I tried to get across in my post, anyway. If the air compressed 25%, then you'd be running around on the rims. If measured accurately enough, the pressures in your snow tire test should not as big of a difference as a few psi.

I did not say that a volume increase resulted in less compression of air.

---jps

 

now that is something i can understand. that makes complete sense thank you.

 

and now we know that they will be 18" x 8"

 

Guess I'll have to set aside some money for tires... or morgage something...

 

No kidding.. 18" high performance summer tires are about $400 Cnd each.. Yikes.. Maybe I can ask for 16's or just order it without the wheels and buy my own 17's. :D

 

I would prefer that the 18" are made part of a 'sport' group (225/45 18 front and 255/40 18 rear on 7.5 and 9.0 inch widths) and 17's with 225/50 17 all around are standard on the RX-8.(on 7 inch widths)

The benefits of symmetric tires are ease of rotation (not that a 50/50 rear axle drive car will (should) have very different tire wear rates)

 

Why do you want to stagger widths? What do you think that would accomplish?

Without even considering replacement cost, I'm looking at the very low profile tires on the 18 inch wheels (look at some of the recent "spy" shots), and I'm getting nervous. I want a little "give" in the tire, so that the rubber can keep a decent grip on the street surface with the imperfect roads we have.

Not that I'm saying it won't work simply due to a couple of pictures that I've seen. I'm just getting a little nervous.

---jps