HW 8 Solution

# HW 8 Solution - FIGURE 11-6 f c 1 ∆ C C Diagram Showing Center Change for Problem 11-6 2 In this case f c 1 0.07 = f c 1.07 = 3 Determine the

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Unformatted text preview: FIGURE 11-6 f c 1 ∆ C C + Diagram Showing Center Change for Problem 11-6 2. In this case, f c 1 0.07 + = f c 1.07 = 3. Determine the pitch diameter and pitch radius of the pinion. Pitch diameter d p N p p d = d p 4.500 in = Pitch radius r p 0.5 d p ⋅ = r p 2.250 in = 4. When the centers are moved apart the base circle diameters don't change but new pitch circle diameters are defined by the intersection of the new line of action, which is always tangent to the two base circles, with the line of centers. The new pitch radii are proportional to f c . Thus, from Figure 11-6, the new pressure angle is cos φ new ( 29 r bp f c r p ⋅ Substituting r p cos φ ( 29 ⋅ for r bp and solving for φ new , φ new acos r p cos φ ( 29 ⋅ f c r p ⋅ = φ new 28.57 deg = PROBLEM 11-6 Statement: What will the pressure angle be if the center distance of the spur gearset in Problem 11-4 is increased by 7%? Given: Tooth numbers: pinion N p 27 = gear N g 78 = Diametral pitch p d 6 in 1- ⋅ = Pressure angle φ 20 deg ⋅ = Solution: See Figure 11-6 and Mathcad file P1106....
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## This note was uploaded on 07/23/2011 for the course EML 4501 taught by Professor Staff during the Spring '11 term at UNF.

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HW 8 Solution - FIGURE 11-6 f c 1 ∆ C C Diagram Showing Center Change for Problem 11-6 2 In this case f c 1 0.07 = f c 1.07 = 3 Determine the

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