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Circular Pads. Fazekas [8.3] has derived the basic equations for circular pads. The effective friction radius is Re = e (8.66)

where e = radius from the center of the disk to the center of the pad and = multiplier found in Table 8.9. Also tabulated in Table 8.9 is the ratio of maximum to average contact pressure.

Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright 2004 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website.

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The actuating force on each pad is F = R2pav (8.67)

Wow, that was easy. Let s see how this can affect the thruster particles:

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The torque capacity per pad is found by using Eq. (8.60) after F and Re have been calculated. Example 9. A sports car requires disk brakes for the front wheels. It has been decided to use two annular pads per wheel with Ri = 3.875 in, Ro = 5.5 in, and = 108 . The friction material supplier guarantees a coefficient of friction of at least f = 0.37. Each pad is actuated by two hydraulic cylinders, each 1.5 in in diameter. Each frontwheel brake provides a braking torque capacity of 13 103 lb in. What hydraulic pressure is needed at the wheel cylinders What are the average and the maximum contact pressures Assume uniform wear. Solution 1. The torque capacity per pad has to be (13 103)/2 = 6500 lb in. Given uniform wear, the effective friction radius is, by Eq. (8.65), Re = R i + Ro 3.875 + 5.5 = = 4.69 in 2 2

The corresponding actuating force is, by Eq. (8.60), T= or F = 1875 lb per wheel cylinder The hydraulic pressure at the wheel cylinder has to be ph = 4(1875) F 4F = = = 1060 psi A p d 2 (1.5)2 p T 6500 = = 3750 lb per pad f Re 0.37(4.69)

2. Equation (8.62) can be used to find the maximum contact pressure: pmax = 3750 F = = 316 psi Ri (R o Ri ) (108 /57.296)(3.875)(5.5 3.875)

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particles.force = BBPointMake(0.0, -20.0, 0.0);

For finding the average contact pressure, use Eq. (8.63): pav = pmax 2Ri /Ro 2(3.875)/5.5 = 316 = 261 psi 1 + R i /Ro 1 + 3.875/5.5

Electromagnetic forces are used in a variety of ways to couple two sides of a clutch or a brake. The use of electrically generated forces implies relatively easy means of automatic control. 8.8.1 Magnetically Actuated Friction Clutches and Brakes The most common use of electromagnetic forces is to provide the actuating force for a friction brake or clutch. The configuration shown in Fig. 8.27 can be used either as a clutch or as a brake. When power is applied to the coil, the magnet, faced with a friction material, attracts the armature. Torque can be varied by using a potentiometer. If one member is fixed, the device functions as a brake; otherwise, it is a clutch. The combination clutch-brake in Fig. 8.28 uses a single solenoid coil. When the coil is deenergized, the clutch is disengaged and springs prevent the release of the brake. When the coil is energized, the clutch is engaged and the brake is released. An arrangement of opposing mechanical springs ensures that the input-side clutch is fully engaged for a brief time before the brake is released. This is done by making the springs assisting clutch engagement weaker than those resisting brake release.

FIGURE 8.27 Electromagnetic friction clutches and brakes. (Warner Electric Brake and Clutch Co.)

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You just need to add this line to your SceneOne thrusters, and they will all get a constant negative y acceleration, which may or may not look like gravity. In Figure 1-11, I turned off the other two emitters so that I could easily see the effect of the force on the top emitter.

Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright 2004 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website.

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