I have made a crane for my engineering project, it has a gearbox with the gear train ratio of 1:4 1:2 1:3 1:4I think the velocity ratio of this is: 1/4 x 1/2 x 1/3 x 1/4 which 0.0104The crane has to lift 1kg, so would 1kg be the effort, or what? and im not clear on any of the formulas, could someone please explain how I would work everything out?
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Bring dried beans with you, and fallow the directions on the back, and you can always add some onions and seasonings to them to make them taste much better.
A basic principle in mechanical engineering is that Work force distance or Power force velocitypower for a rotating shaft is: Power torque angular velocity How does this apply? Well, you got the velocity ratio correctThe theoretical mechanical advantage is just 1/ratio, or 4x2x3x4 or 96Now, the velocity ratio will always be correct, because the gears have teeth, and force the speeds to the correct ratios, or 1/96Note that the power is 1/96 96, or 1x the original powerYou have one other issue thoughThere is one more stage of reduction: where the rope winds on the spindleForce dist force (radius of spindle + 1/2 dia of rope)Its a little complicated, probably best to leave it out! In your gear train, you have frictionThis loses power, and the true mechanical advantage is the force out/force inThis is tough to measure1kgf is the effort to lift the weight, but you can't tell how much torque the motor is generatingBest just to say that unlubricated gear trains typically lose 5% per stage, thus, efficiency (95%)^4, or about 81% efficientYou also have friction in your motor, as well as some waste because it is not perfect in converting electrical energy into mechanical energyEfficiency is power out/power inIf you can measure the voltage on the motor, the current in the motor, and the time to lift a certain distance, you can calcuate efficiency as below: Energy in volt?amp?time Energy out 1 kg?9.8m/s??distance[in m] Both are in Joule/sec.
