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Figure 4.81 shows two angled faces that have to be tested for correctness with the aid of rollers, 15 mm in diameter. Calculate the distance M between the rollers for the angles to be correctly machined.

Mathematics

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Answer

52.80 mm52.80 \text{ mm}

Step-by-step solution

6 steps

To calculate the distance MM between the rollers, we need to determine the horizontal position of the center of each roller relative to the corners of the 80 mm80 \text{ mm} base.

  1. Identify the given values and calculate the roller radius.

    The diameter of the rollers is d=15 mmd = 15 \text{ mm}.
    The radius of each roller is r=d2=152=7.5 mmr = \frac{d}{2} = \frac{15}{2} = 7.5 \text{ mm}.
    The bottom width of the groove is 80 mm80 \text{ mm}.

  2. Determine the angle of the left inclined face with the horizontal.

    The diagram shows an internal angle of 130∘130^\circ between the top horizontal line and the left inclined face.
    The angle of the left inclined face with the horizontal, θL\theta_L, is 180∘−130∘=50∘180^\circ - 130^\circ = 50^\circ.

  3. Calculate the horizontal distance from the left corner to the center of the left roller.

    For a roller tangent to a horizontal surface and an inclined surface, the horizontal distance from the corner (where the two surfaces meet) to the center of the roller is given by x=rcot⁡(θ/2)x = r \cot(\theta/2), where θ\theta is the angle of the inclined surface with the horizontal.
    For the left roller:

    x1=rcot⁡(θL2)=7.5cot⁡(50∘2)=7.5cot⁡(25∘)x_1 = r \cot\left(\frac{\theta_L}{2}\right) = 7.5 \cot\left(\frac{50^\circ}{2}\right) = 7.5 \cot(25^\circ) x1=7.5×1tan⁡(25∘)x_1 = 7.5 \times \frac{1}{\tan(25^\circ)} x1≈7.5×10.466307658x_1 \approx 7.5 \times \frac{1}{0.466307658} x1≈7.5×2.14450692x_1 \approx 7.5 \times 2.14450692 x1≈16.0838 mmx_1 \approx 16.0838 \text{ mm}
  4. Determine the angle of the right inclined face with the horizontal.

    The diagram shows an internal angle of 112∘112^\circ between the top horizontal line and the right inclined face.
    The angle of the right inclined face with the horizontal, θR\theta_R, is 180∘−112∘=68∘180^\circ - 112^\circ = 68^\circ.

  5. Calculate the horizontal distance from the right corner to the center of the right roller.

    For the right roller:

    x2=rcot⁡(θR2)=7.5cot⁡(68∘2)=7.5cot⁡(34∘)x_2 = r \cot\left(\frac{\theta_R}{2}\right) = 7.5 \cot\left(\frac{68^\circ}{2}\right) = 7.5 \cot(34^\circ) x2=7.5×1tan⁡(34∘)x_2 = 7.5 \times \frac{1}{\tan(34^\circ)} x2≈7.5×10.674508516x_2 \approx 7.5 \times \frac{1}{0.674508516} x2≈7.5×1.48256091x_2 \approx 7.5 \times 1.48256091 x2≈11.1192 mmx_2 \approx 11.1192 \text{ mm}
  6. Calculate the distance MM.

    The distance MM is the horizontal distance between the centers of the rollers. This can be found by subtracting the horizontal distances x1x_1 and x2x_2 from the total bottom width of 80 mm80 \text{ mm}.

    M=80−x1−x2M = 80 - x_1 - x_2 M=80−16.0838−11.1192M = 80 - 16.0838 - 11.1192 M=80−27.2030M = 80 - 27.2030 M=52.7970 mmM = 52.7970 \text{ mm}

    Rounding to two decimal places:

    M≈52.80 mmM \approx 52.80 \text{ mm}

    The final answer is 52.80 mm52.80 \text{ mm}.

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