Question

Im designing a boat anchor winch that anchors a load of 1000 kg into the water when the boat is stationary with a wire line from the winch to the anchor, I came up with a spur gear design a pinion gear 1 and gear 2. Gears is made of carbon steel. For that application what is the correct motor to connect into the spur gear. How much horseower does gear 1 and gear 2 need to operate a 1000 kg load. You can estamate the rest of the variables.

Answer 1

Since the load is 1000 Kg, assuming the boat to be sufficiently large to hold a 60 cm diameter gear.

Following are the assumptions:

1. Gear has a diameter of 60 cm or 0.6 m.

2. Pinion has a diameter of 20 cm or 0.2 m.

3. During operation, the load is being raised/lowered at a speed of 60 cm/sec or 0.6 m/sec.

4. Mechanical efficiency is 100%.

5. The load is so designed that the hydrostatic force by water can be neglected.

We know that,

Power, P = T x

Also, power remains constant.

So, T₁₁ = T₂₂

where subscripts 1 and 2 stand for pinion and gear respectively.

V₂ = R₂₂

0.6 = 0.3 x ₂

Therefore, ₂ = 2 rad/sec.

T₂ = (Load) x R₂

= 1000 x 9.81 x 0.3

= 2943 Nm.

Therefore, Power = 2943 x 2

= 5886 Nm/s

= 5.86 KW.

1 HP = 746 W

Therefore, power (in HP) = 7.89 HP.

Standard motor power = 8 HP

Note: This power is calculated based on 100% mechanical efficiency meaning friction power is neglected. For more accurate results, factors such as form and finish of gears, co-efficient of friction, module, etc. should also be considered.

Note: You can have a considerably lower power motor by reducing the speed with which the load is being raised/lowered.

Torque on motor = 1000 x 9.81 x (0.2/2)

= 981 Nm

T₁₁ = T₂₂

Therefore, ₁ = (2943 x 2)/981

= 6 rad/sec.

Motor speed = 57.2 ~ 58 RPM

Ans: Power 8 HP, Speed 58 RPM