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Parents Pricing Home NSC Mechanical Technology Fitting and Machining Calculations on size of drills for bolts and nuts (ISO metric) 6.1 'n Regtutdrat, met 'n steeksirkeldiameter van 165 mm en 110 tande, word vir 'n rattkas benodig
6.1 'n Regtutdrat, met 'n steeksirkeldiameter van 165 mm en 110 tande, word vir 'n rattkas benodig - NSC Mechanical Technology Fitting and Machining - Question 6 - 2022 - Paper 1 Question 6
View full question 6.1 'n Regtutdrat, met 'n steeksirkeldiameter van 165 mm en 110 tande, word vir 'n rattkas benodig.
Bereken die volgende:
6.1.1 Module.
6.1.2 Buitediameter.
6.2 ... show full transcript
View marking scheme Worked Solution & Example Answer:6.1 'n Regtutdrat, met 'n steeksirkeldiameter van 165 mm en 110 tande, word vir 'n rattkas benodig - NSC Mechanical Technology Fitting and Machining - Question 6 - 2022 - Paper 1
6.1.1 Module. Only available for registered users.
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To calculate the module (SSD), we can use the formula:
ext{Module} = rac{ ext{SSD}}{T} = rac{165}{110} = 1,5
Thus, the module is 1,5.
6.1.2 Buitediameter. Only available for registered users.
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The outer diameter (BD) can be calculated using the formula:
e x t B D = e x t S S D + 2 ( m ) = 165 + 2 ( 1 , 5 ) = 168 e x t m m ext{BD} = ext{SSD} + 2(m) = 165 + 2(1,5) = 168 ext{ mm} e x t B D = e x t SS D + 2 ( m ) = 165 + 2 ( 1 , 5 ) = 168 e x t mm
Therefore, the outer diameter is 168 mm.
6.2.1 Maksimum wyte (W) van die swaelstert. Only available for registered users.
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To find the maximum width (W) of the counter, we use:
W = 120 + 2 ( D E ) W = 120 + 2(DE) W = 120 + 2 ( D E )
If we calculate DE:
D E = a n 3 0 e x t ° i m e s A D = 17 , 32 e x t m m DE = an 30^ ext{°} imes AD = 17,32 ext{ mm} D E = an 3 0 e x t ° im es A D = 17 , 32 e x t mm
Thus, substituting DE:
W = 120 + 2 ( 17 , 32 ) = 154 , 64 e x t m m W = 120 + 2(17,32) = 154,64 ext{ mm} W = 120 + 2 ( 17 , 32 ) = 154 , 64 e x t mm
6.2.2 Afstand (m) tussen die presisirollers. Only available for registered users.
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To find the distance (m) between the rollers, we calculate:
m = W − 2 ( A C ) − 2 ( R ) m = W - 2(AC) - 2(R) m = W − 2 ( A C ) − 2 ( R )
Assuming we have calculated AC as:
A C = 19 , 05 e x t m m AC = 19,05 ext{ mm} A C = 19 , 05 e x t mm
Then substituting:
m = 154 , 64 − ( 2 ( 19 , 05 ) + 2 ( 11 ) ) = 94 , 54 e x t m m m = 154,64 - (2(19,05) + 2(11)) = 94,54 ext{ mm} m = 154 , 64 − ( 2 ( 19 , 05 ) + 2 ( 11 )) = 94 , 54 e x t mm
6.3.1 Bereken die indeksering wat benodig word. Only available for registered users.
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The scoring can be calculated by:
ext{Indekering} = rac{n}{40} = rac{40}{163}
Thus, simplifying gives:
ext{Indekering} = rac{6}{24}
6.3.2 Bereken die wisselratte wat benodig word. Only available for registered users.
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Using the formula for the gear ratio:
G_d = rac{(A - n) imes 40}{A}
Substituting A yields:
G_d = -rac{3 imes 40}{160} = -120
6.4 Noem TWEE tipe balanseermetodes. Only available for registered users.
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Statiese balanseering (still balancing).
Dinamiese balanseering (running balancing).
6.5 Noem TWEE voordeel van korrekte balasering. Only available for registered users.
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