... I'm looking at 5 amp 12v power supplies, do you think that would be enough?...
If the power supply is well regulated, that should do the job nicely. By "well regulated" I mean the terminal voltage of the power supply does not change significantly from a nominal 12 V DC as the current demand increases from zero to its rated value of 5 A. You can check this with a load resistance connected to the power supply terminals. By Ohm's law, a resistor drawing 5 A at 12 V will have a value R = V / I = 12 / 5 = 2.4 Ω and will dissipate power P = V I = (12)(5) = 60 watts.
It is unlikely you will have a 2.4 ohm resistor rated for 60 watts handy for testing, and it is not cost-effective to purchase one just to test your power supply. An alternative is to scrounge a few 12 V automobile lamps... a headlamp would be appropriate, but brake lights and turn-signal lights will also work if you connect several in parallel to reach the 5 A maximum current draw. Don't use LED lights; they typically do not draw enough current to adequately "load" your power supply.
Better still, use your wiper motor as the load! Connect it to the power supply
before turning on the power supply (to avoid "arcs and sparks" while connecting wires up), energize the power supply
briefly and observe whether or not the motor runs. If so, you are almost done.
If the motor does not run, quickly turn off the power supply (to avoid overheating and damaging something), disconnect the motor and begin troubleshooting the problem. We will try to help you with that if the motor doesn't run when it is supplied adequate voltage and current.
... "Please tell us what this "short circuit" current is AND the power supply terminal voltage that occurs when this happens." I'm sorry Hevens I don't understand this, I haven't learned how to measure current yet.
The short-circuit current is the current a power supply will supply at zero terminal voltage. If the load is less than a short circuit, there will be some terminal voltage when the load is applied. The two measurements of short-circuit current and terminal voltage (if any) along with the open-circuit voltage can be used to calculate the internal resistance of the power supply. The internal resistance causes the power supply terminal voltage to decrease with increasing current load. A voltage regulated power supply uses negative feedback to maintain a constant and low internal resistance from no load up to full load so there is no appreciable drop in terminal voltage from zero current to full-load current. Switch-mode power supplies are very good at this and more efficient than linearly-regulated power supplies.
I am going to have to resort to the dreaded
water analogy to describe voltage and current measurements. Like any analogy, it does not describe what is really going on, but it can be useful for describing concepts in a more intuitive manner.
In a water delivery system, whether public or private well, water is delivered through pipes by a pump that pressurizes the water in the pipes. Now everyone knows water is pretty much not compressible, so what does this mean? It simply means that there is a force (provided by the pump) that will move the water through the pipes. Pressure is force per unit area. It does not mean that the substance to which the pressure is applied actually responds by compressing, i.e., reducing in volume or increasing in density. So the pump pressurizes the water and this force is measured in pounds per square inch (psi) or more specifically in gauge pressure (psig) meaning the pressure is measured relative to the atmospheric pressure on the outside of the gauge. In this analogy, pressure is equivalent to the voltage in an electrical circuit.
When you turn on a spigot to draw water from the pipes, the water flows through the spigot at a rate you can measure in gallons per minute. In this analogy, water flow rate is equivalent to the current in an electrical circuit. The rate of flow is determined by, among other things, the size of the spigot and the water pressure. The size of the spigot is analogous to resistance in an electrical circuit. Larger spigots have less resistance to water flow than small spigots. Higher pressure means more water flow. To measure this flow you need to insert a flow-meter between the pressurized water source and the spigot. To measure current in an electrical circuit, you need to insert a current-meter (ammeter) between the voltage source and its load.
At this point I will discard the water analogy. I only used it to demonstrate the difference in how you measure voltage (water pressure) and current (water flow) in an electrical circuit. Water flow and electrical flow are not the same phenomena, and the physics and math for water flow are way more complicated than the simple Ohm's Law we use to calculate electrical voltages, currents, and resistances.
Okay, I will assume that you
do know how to use an inexpensive digital multi-meter to measure AC and DC voltages, resistance, continuity, and currents up to a few hundred milliamperes (100 mA = 0,1 A). It will typically have only two test leads, black for the
common test lead and red for the
input test lead. The two leads are always used together as a
pair of test leads. There is more you should learn, but simplistically, a modern digital multi-meter is just a liquid crystal display screen, a function switch, and two test probes. If you don't understand that, please let us know right away. If you don't own a digital multi-meter, purchase one right away.
Most multi-meters have a measuring function for high currents that is completely separate from their other functions of measuring volts, ohms, and milliamperes (VOM). Typically there is only one such range and it covers zero to 10 A or zero to 20 A, depending on the meter manufacturer. A common
test lead (usually black in color) is used for all mult-imeter measurements, along with another
test lead (usually red in color) for measuring voltages with respect to the common lead, resistance in ohms between the two leads, continuity (low resistance) between the two leads, and milliampere currents between the two leads. However, for high currents, another separate jack is provided on the meter case for the multimeter
test lead that is used with the common
test lead.
To measure current, you have to place the two meter leads
in series with the load and set the meter range/function switch to the position used to measure high current. That means you have to disconnect one terminal of the load from the power supply. You now have two terminals in front of you: one on the load, one on the power supply. Connect the meter between those two terminals, making sure it is set to the high-current measuring range and the
test leads are plugged into the high-current test jack and the common jack. With digital multi-meters polarity doesn't matter because the meter will always display the polarity. One way will read positive current, and if you reverse the leads to the meter, that other way will read negative current. It's the same current, either way, so ignore the polarity indication.
Until you get more experience, always make the current-metering connections with power off. Turn the power on and measure the current after the connections to do so are established.
Another pitfall to avoid is forgetting the test leads are hooked up for measuring current, and then moving the red test lead to a point where you want to measure voltage. This will usually result in a short circuit that blows a fuse inside the multi-meter. That's right up there with measuring the resistance in a circuit and then trying to measure a voltage without changing the function switch. Both nubie mistakes can result in damage to the multi-meter.
.. I intend on using the motor as in a dead lift scenario, 4cm cam off motor axle, lifting around 1.5kg off an arm around 30cm long, so it will be fairly heavily loaded. Not sure how to calculate this into amps needed, I'm in the middle of learning torque and torque conversion calculations at the minute, getting there slowly.
This is probably NOT going to work. The motor shaft will turn too quickly (hundreds if not thousands of RPM) to allow control of the cam. There IS a way to make it work by using pulse-width modulation to control the motor speed, but you are still looking at huge torque requirements to lift 1.5 kg at the end of a 30 cm lever arm with a 4 cm cam, even if the cam is located directly under the load. You don't specify what the total cam "lift" will be, or where that "lift" is applied to the 30 cm arm. A clean sketch using a black felt-tip marker on plain white paper, well lighted with no shadows, should be photographed and uploaded for our inspection and comments.
I fear you are going to need a geared motor for your lifting application. The windshield wiper gear box contains a worm gear driving a spur gear that in turn drives the reciprocating mechanism. The motor always runs in one direction, at two speeds, and the reciprocating mechanism may not appreciate being driven in reverse. It may jam if reversed. If possible, open the gear box and do your "power take-off" from the shaft of the main gear that is driven by the worm gear. Discard everything else. Or purchase a
worm gear speed reducer box and adapt it to fit your motor. Or purchase a
geared motor as others here have suggested. But do the torque and speed calculations first, then decide what you need or can adapt to your purpose.
Is your contraption gonna automagically make a musical beat for accompanyment?
