Author: Site Editor Publish Time: 17-10-2025 Origin: Site
The output speed of the SA47-15-1.5KW worm reducer is determined by three core factors: 'transmission ratio', 'motor input speed' and 'actual working condition loss'. Among them, the transmission ratio and motor input speed are the basis for theoretical calculation, and the working condition loss is a correction term in actual operation. The following is a disassembly analysis of the specific influencing factors, combined with the model parameters ('SA47' is the frame number, '15' is the nominal transmission ratio, '1.5KW' is the matching motor power) to expand the explanation:
1. Core determinant: transmission ratio ('fixed coefficient' of theoretical output speed)
The essence of the output speed of the worm reducer is 'motor input speed ÷ transmission ratio', where the transmission ratio is fixed and determined by the mechanical structure of the reducer (number of worm heads, number of worm gear teeth), and is the core and most stable factor that affects the output speed.
1. Definition and calculation logic of transmission ratio
The transmission ratio (i) is the ratio of the input speed (n₁) and the output speed (n₂) of the reducer. The formula is:
i = n₁ ÷ n₂ → n₂ = n₁ ÷ i
For SA47-15-1.5KW, '15' in the model refers to the nominal transmission ratio (i=15), which means that for every 15 turns of the motor, the output shaft of the reducer turns 1 turn.
2. The 'structural source' of the transmission ratio (why is it a fixed value?)
The transmission ratio is directly determined by the 'gear ratio' of the worm and worm gear. The specific formula is:
i = Number of worm gear teeth (z₂) ÷ Number of worm heads (z₁)
Number of worm heads (z₁): That is, the number of spirals of the worm (commonly 1, 2, or 4 heads). The more heads, the higher the transmission efficiency, but the smaller the transmission ratio;
The number of worm gear teeth (z₂): matches the number of worm heads. The more teeth, the larger the transmission ratio, but the size of the worm gear will increase accordingly (limited by the space of the SA47 machine base, the tooth number range is fixed).
Taking SA47-15 as an example, if the matching worm is '1 head', then the number of worm gear teeth = 15×1=15 teeth; if it is '2 heads', then the number of worm gear teeth = 15×2=30 teeth - regardless of the number of heads, the number of teeth of the worm and worm gear is fixed at the factory, so the transmission ratio i=15 is an 'unchangeable mechanical attribute', which directly determines the theoretical basis of the output speed.
2. Basic input factors: rated input speed of supporting motor
The theoretical value of the output speed needs to be based on the 'motor input speed'. Therefore, the number of poles and frequency (power supply parameters) of the motor directly affect the input speed, and then indirectly determine the output speed.
1. The relationship between the number of motor poles and the rated speed
The matching motor of SA47-15-1.5KW is a 'three-phase asynchronous motor', and its rated speed is determined by the 'number of motor poles' (the number of poles is the number of magnetic pole pairs of the motor stator winding, commonly 2, 4, 6, and 8 poles), the formula is:
Motor rated speed (n₁) ≈ 60 × power frequency (f) ÷ number of motor pole pairs (p) × (1 - slip s)
Power frequency (f): my country's industrial electricity standard is 50Hz (some countries are 60Hz), which is a fixed value;
Slip (s): The inherent characteristics of asynchronous motors. Under rated load, s≈0.02~0.05 (that is, the speed is 2%~5% lower than the 'synchronous speed'), which can be approximately ignored.
2. Influence of power frequency (special scenario)
If the reducer is used in a non-50Hz power supply scenario (such as export equipment using 60Hz power supply), the synchronous speed of the motor will increase, which will cause the output speed to increase. For example:
Under 60Hz power supply, the synchronous speed of the 4-pole motor = 60×60÷2=1800r/min, the rated speed ≈1746r/min, and the output speed of the reducer ≈1746÷15≈116.4r/min (about 19.4r/min higher than that at 50Hz).
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