Wheel Dimensioning Game

Many and many people will be misunderstood when choosing the hub size, especially if they cannot systematically calculate their own size. Now I want to write this essay to share with you the experience and methods of my personal wheel hub, hoping to help everyone.

What are the basic parameters of the hub, often can make many people worry about. Many people will confuse the size of the hub with the size of the tire when talking about the hub size. For example, A asked B: “How wide is your hub?” B replied: “My hub 18 has a width of 245”. Here is the most obvious concept confusion: "245" is the width of the tire rather than the width of the hub. How to calculate the size of the tire and read it later I will mention, now we first focus on the size of the hub.

The correct notation for the size of the hub is: AA X BB J+/-CC

"AA" is the diameter of the hub. That is, the number of wheels that are commonly referred to as 18-inch wheels and 20-inch wheels.

"BB" is the width of the hub, and the back "J" is always associated with the previous value, which is an inch. Now implant a concept for a better explanation later, 1J = 25.4 mm.

"CC" is the negative value of the hub, also known as ET, in millimeters. It is a very important but not entirely decisive factor for the overall position of the hub.

The "X" in the middle of "AA" and "BB" is a multiplication sign, read as "xx multiplied by xx J"

Understand the three elements of the hub size, for example:

The size of a set of wheels is written as 19X10J+35. Then we know that this is a 19-inch wheel with a width of 10 inches and a negative value of 35mm.

How can these values ​​help us choose the hub? The two most important elements: width and negative value.

Let's talk about width first. The hub width is "BB" J, which is a number multiplied by a "J". As mentioned earlier, 1J is equal to 25.4 mm so the width of the hub is equal to a number times the length of 25.4 mm, and the final unit is mm. However, it should be noted that when the width of the hub changes, the hub is uniformly widened or narrowed as a whole. So the change in width is reflected in the average of both sides of the hub. That is to say, let's just leave do not say that the 18x10J hub is wider than the 18x9J hub (assuming that the et is the same) by 0.5 J on each side of the hub. The principle of this operation is 10J-9J=1J, (the width difference between the two hubs) 1J/2=0.5 J (the width difference between the two sides of the hub). Finally, substituting the concept of 1J=25.4 mm, we know that one side is more than before. 12.7 mm.

So suppose you have a set of wheels with a size of 18x10.5J+10 as your template. You want to choose a set of wheels to keep the same, but to be 25.4mm outside this set of wheels, then you need to have 1J on the outside, which is equal to the overall ratio. This set of template wheels is 2J more. The final size you want is 18x12.5J+10.

Let's talk about the negative value et. The most important concept here is that the smaller the negative value, the closer the hub is to the exterior/fender. So do not say aside the width of the hub, et10 is a little more violent than et15. So suppose that the current template hub size is 17x8J+35. You now want to change the hub to keep the width constant, but it will be about 14mm more into the body. Then let's calculate: Keep the width unchanged, then 17x8J copy in. However, if the hub is retracted into the body, it is to increase the negative value, +35+14=+49. The final size we want is 17x8J+49.

Having said two basic concepts, we can begin to calculate the hub size and choose the hub we want. for example:

Now that I know the location of a set of 18x8.5J+28 wheels in the car, after estimating and measuring, I need to move closer to the outside of the car body by about 38mm. So I have three ways to achieve:

1, negative value unchanged: 38.1mm = 25.4mm + 12.7mm = 1J + 0.5J Then I know that I need more than one unilateral 1.5J, overall more than 3J. Results: 18x11.5J+28

2, the same width: 18x8.5J + 28 data directly minus 38 mm - 18x8.5J (+28-38) = final data 18x8.5J-10

3, both change: the reality is cruel, the general wheel in the actual production are both negative and the width of the change at the same time, the width of the same or an ideal state of negative value is often not the case, because the violation of the wheel hub The physical limitations of design and manufacturing: If the negative value is too high, there is no way to avoid the brake. If the value is too low, the hub is too violent. If the width is too small, the brakes cannot be installed on the tires. So both are going to change under the rational range is really should be the method we need to take when choosing the hub. Because width is one of the most important factors, because it determines the range of negative values ​​and the choice of tires, so we must first understand how wide the hub needs. For example, I need a wider wheel not only for tire selection but also for a wider wheel. Therefore, we initially determined the hub width was selected as 10J, so we got initial data of 18x10J. So come to a simple calculation about 10J more than 8.5J 1.5J. 1.5J = 1.5 x 25.4 = 38.1 mm. Then divide by 2 to get 19.05 millimeters wider than before. So assuming the same negative value, 18x10J is 19.05 mm more out than 18x8.5J. Now let's introduce the concept of negative values. Remember that before I said that the smaller the negative value is, the closer the hub is to the outside, then when the width already has a basic line (more than 19.05mm outside), we are still at a distance of 38mm from the ideal target. It is corrected by a negative value. 38-19.05 = 18.95 mm The final original negative value is +28-18.95 = +9.05 mm which is equal to 9 mm, so our final ideal size is 18x10J + 9

After we understand how to calculate, this size can be freely changed, and the same position size is 18x9.5J+3 and 18x10.5J+15.

Finally, let me stress one more point. When discussing the loading position and size of the hub, it is totally unconvincing to look at only one of the data between the width and the negative value. The combination of the two must be considered. Therefore, the two are indispensable and one factor is missing. The other data becomes completely useless.