14 GEARS March 2006 I n the past editions of Basic Hydraulics, we’ve covered how various types of pumps function and how the pressure is regulated. In this edition we’ll cover how this pres- sure is used and the difference between pressure and force. We’ll also show you how to calculate the amount of force, as long as you know the pressure and what the area is that this pressure is being applied to. But why is this important? It’s important because with this infor- mation, you can calculate line pressure, accumulator pressure, shift timing, clamping force of a particular clutch pack or band servo and a lot more. What is pressure? In hydraulics, it can be defined as force applied uniformly over a contained area and is measured in force per unit of area. The most common way in our industry to measure hydraulic pressure is in PSI (pounds per square inch). If pressure is contained in an area like in a clutch drum, pressure is equal anywhere in that area that you measure it. Sound confusing? Lets look at some examples of areas first. Figure one illustrates a scale with a 1sq. in. apply surface. If you were to apply 70 psi of pressure to the surface, there would be 70 lbs of force on the scale. If you were to apply 100 psi to the surface there would be 100 lbs of force on the scale. If you were to increase the apply area to 9 sq in, and applied 70 psi of pressure to this larger area, you would have 630 lbs of force to the scale (figure 2). This is because at 70 psi there is 70 pounds of force applied to each of the 9 square inch (70 x 9 = 630). That part is fairly simple to under- stand. But how many square apply areas do you see in a transmission? The trickiest part about calculating apply areas is that the parts are always round! There are two different calculations that can be used to calculate the area of a circle. One is the traditional and most accurate…also very difficult to use. And the other is much easier to use but a little less accurate. The difference in the accuracy is so small however that it really makes no measurable difference. πr² = The Area of a Circle This is the traditional calculation. The “r” symbolizes radius which is half of the diameter. The “π” symbol- izes pi which is the circumference of a one inch diameter circle. So to use this formula to calculate the area of a circle, you first you have to divide the diameter by 2 then multiply that by BASIC HYDRAULICS In hydraulics, it can be defined as force applied uniformly over a contained area and is measured in force per unit of area. What is What is Pressure? Pressure? by Larry Frash Figure 1 Figure 2 70 psi x 1 sq. in. = 70 lbs of force 70 psi x 9 sq. in. = 630 lbs of force
Transcript
14 GEARS March 2006
In the past editions of Basic Hydraulics, we’ve covered how various types of pumps function
and how the pressure is regulated. In this edition we’ll cover how this pres-sure is used and the difference between pressure and force. We’ll also show you how to calculate the amount of force, as long as you know the pressure and what the area is that this pressure is being applied to. But why is this important? It’s important because with this infor-mation, you can calculate line pressure, accumulator pressure, shift timing, clamping force of a particular clutch pack or band servo and a lot more.
What is pressure? In hydraulics, it can be defined as force applied uniformly over a contained area and is measured in force per unit of area. The most common way in our industry to measure hydraulic pressure is in PSI
(pounds per square inch). If pressure is contained in an area like in a clutch drum, pressure is equal anywhere in that area that you measure it. Sound confusing? Lets look at some examples of areas first.
Figure one illustrates a scale with a 1sq. in. apply surface. If you were to apply 70 psi of pressure to the surface, there would be 70 lbs of force on the scale. If you were to apply 100 psi to the surface there would be 100 lbs of force on the scale. If you were to increase the apply area to 9 sq in, and applied 70 psi of pressure to this larger area, you would have 630 lbs of force to the scale (figure 2). This is because at 70 psi there is 70 pounds of force applied to each of the 9 square inch (70 x 9 = 630).
That part is fairly simple to under-stand. But how many square apply
areas do you see in a transmission? The trickiest part about calculating apply areas is that the parts are always round! There are two different calculations that can be used to calculate the area of a circle. One is the traditional and most accurate…also very difficult to use. And the other is much easier to use but a little less accurate. The difference in the accuracy is so small however that it really makes no measurable difference.
πr² = The Area of a Circle
This is the traditional calculation. The “r” symbolizes radius which is half of the diameter. The “π” symbol-izes pi which is the circumference of a one inch diameter circle. So to use this formula to calculate the area of a circle, you first you have to divide the diameter by 2 then multiply that by
BASIC HYDRAULICS
In hydraulics, it can be defined as force applied uniformly over a contained area and is measured in force per unit of area.
What isWhat is Pressure?Pressure?by Larry Frash
Figure 1 Figure 2
70 psi x 1 sq. in. = 70 lbs of force 70 psi x 9 sq. in. = 630 lbs of force
basicHYD4-06.indd 14basicHYD4-06.indd 14 3/6/06 10:50:39 AM3/6/06 10:50:39 AM
GEARS March 2006 15
itself. Then, multiply that number by pi (3.14159265). So to calculate the area of a 4 inch circle (figure 3), the equa-tion would look something like this. 2x2x3.14159265=12.5663706. One more thing to point out is that pi never ends, it has to be rounded off.
DxDx0.7854 = The area of a circle
This calculation is much more simple and is accurate enough for any calculations we would do in our indus-try. To use this formula to calculate the area of a circle all you have to do is multiply the diameter x diameter x
0.7854. To calculate a 4 inch circle, the equation would look like this, 4 x 4 x 0.7854 = 12.5664. If you compare this answer to the traditional way, you’ll see that it is very close; especially when you consider rounding the number to the 100th decimal. The answer from both of these equations would both equal 12.57 sq in.
Figure 3 Figure 4
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70 psi x 12.57 sq. in. = 879.9 lbs of force
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16 GEARS March 2006
If you were to put this 4” diameter piston on the scale with 70 psi pressure added to it, the force would be 879.91 lbs (figure 4).
But what about pistons that have a hole in the center. The key is that you first have to calculate the area of the unused area and subtract it from the total area of the piston. Notice I said the word area. That’s because if you subtracted the diameters instead of the areas, the math won’t work.
The piston in figure 5 has a total diameter of 6 inches which calculates to a total area of 28.27 sq. in. (6 x 6 x 0.7854). The unused portion of this piston has a diameter of 2 inches which calculates to an area of 3.14 sq in (2 x 2 x 0.7854). When you subtract the
unused area from the total area, the used area equals 25.13 sq in (28.27 – 3.14). If you put this piston on the scale at 70 psi, the apply force would equal to 1759.1 lbs (figure 6). Enough to lift a small car! Imagine the force on this piston if the mainline pressure exceeded 300 psi. Well, don’t imag-ine…just do the math.
Now let’s look at a dual area piston. This piston is similar to one you would find in a 4L80E direct clutch drum. There are two separate apply areas that are used. To calculate the working or used areas (light blue), you must again subtract the unused area from the total area. See examples (figure 7 & 8). The large outer area would have a working area of 15.70 sq in. The small inner area
What is Pressure?
Figure 5 Figure 6
Figure 7 Figure 8
There are two separate apply
areas that are used. To calculate the working or used
areas (light blue), you must again
subtract the unused area from the total area. See examples
(figure 7 & 8).
6 x 6 x 0.7854 = 28.27 sq. in. 2 x 2 x 0.7854 = 3.14 sq. in. working area = 25.13 sq. in.
6 x 6 x 0.7854 = 28.27 sq. in. 4 x 4 x 0.7854 = 12.57 sq. in. working area = 15.70 sq. in.
4 x 4 x 0.7854 = 12.57 sq. in. 2 x 2 x 0.7854 = 3.14 sq. in. working area = 9.43 sq. in.
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would have a working are of 9.43 sq in. A 4L80E in reverse uses both apply areas in reverse so you would add these area together for a total apply area of 25.13 sq in. At 280 psi line pressure, the direct clutch in this example would have 7036.4 lbs of apply force!
Servos are a little different to cal-culate but you still subtract the unused area from the total area. The servo in figure 9 is similar to a C-6 in that it uses the inner area of the servo to apply the servo. The servo in this example has an apply diameter of 2 inches which calcu-lates to an apply area of 3.14 sq in.
The servo in figure 10 is similar to a 3T40 in that it uses the outer area to apply the servo. To calculate the apply area you would subtract the unused area
from the total area. The servo in this example has a total area of 12.57 sq in. So, when you subtract 3.14 sq in from the total area, the apply area would be 9.43 sq in.
The difference between clutch pis-tons and servos is that most servos use pressure on the pin side to release the servo. To calculate the release force of a servo, you must subtract the area of the pin from the total area of the servo (figure 11).
As you can see, there’s a lot more about hydraulics than can be covered in one article. So we’ll wait until the next edition of Basic Hydraulics to dig deeper and explain more ways to use hydraulics.
What is Pressure?
Figure 9 Figure 10
Figure 11
Servos are a little different to calculate but you still subtract
the unused area from the total area.
The servo in figure 9 is similar to a C-6 in
that it uses the inner area of the servo to
apply the servo.
Total apply area = 3.14 sq. in. Total area of the 4'' servo = 12.57 sq. in. Unused area = 3.14 sq. in. Total apply area = 9.43 sq. in.
Total area of the 4'' servo = 12.57 sq. in. Pin area = 0.126 sq. in. Total release area = 12.44 sq. in.
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