Motor Constraints
Dec. 4, 2020
This is the second part on designing a face-plate lathe for turning large objects. In the first part, we determined various numbers needed to engineer the basic structure. In this part, the problem of how to drive the spindle is addressed. This turns out to be the major constraint.
In some ways this is the hardest part of the entire project. The lathe needs to be able to turn as slow as 10 RPM for large objects, and as fast as 1000 RPM for smaller objects. That is a huge range, and it's not clear how provide such a wide range.
How much torque and power do we need? Once the object is up to speed, it doesn't take much to counter the loss due to the cutting tool. Leaving aside the friction inherent in a larger machine, and assuming cutting tools of the same size, it doesn't matter whether you're turning an object four inches in diameter or four feet; the loss to the continous cutting action is the same either way, and requires horsepower. Getting a large object up to speed takes torque.
What follows uses metric units almost exclusively since it is so much easier.
Before doing things the right way, using torque, consider the following magical line of reasoning. The spinning object has a certain amount of kinetic energy, and a motor produces power (i.e., energy per unit time). If this power is magically dumped into the kinetic energy of the spinning object, it will take some time for the kinetic energy to reach the desired level. This provides limits on the time/power relationship.
The kinetic energy of a spinning object is given by \[ E = {1\over 2}I\omega^2, \] where \(I\) is the moment of inertia and \(\omega\) is the angular velocity, in rad/sec. The moment of inertia for a spinning disc is \[ I = {1\over 2} Mr^2, \] where \(r\) is the radius and \(M\) is the mass. It's easier to work in rev/sec instead of rad/sec, so write the energy as \[ E = {1\over 4} Mr^2 (2\pi R)^2 = \pi^2 Mr^2 R^2, \] where \(R\) is in rev/sec.
Earlier, we observed that the top linear speed (i.e., at the circumference of the object) will be about \(S = 16\) m/sec (52.5 feet/sec). This speed is given by \[ S = 2\pi rR, \] so that \[ R = {S\over 2\pi r} = {8\over\pi r}. \] Plugging this into the equation for energy, we have \[ E = \pi^2 Mr^2 \left({8\over\pi r}\right)^2 = 64 M, \] which is very nice since it is independent of \(r\). This energy is in Joules (N-m) and a 1 hp motor provides 746 Watts or 746 J/s.
Divide the target energy level, \(64M\) by the rate at which energy is added, to find that the amount of time required to reach a total energy output of \(E\) is \[ t = {64M\over 746},\] in seconds. When turning small objects, weighing only a few kg, \(t\) is well under 1 second. But suppose that \(M=150\) kg. Then \[ t = {64\cdot 150\over 746}\approx 13 {\rm\ sec}.\] We can either be patient or use a larger motor. But even a 5 hp motor would take 2.57 sec to spin up an object weighing 150 kg — and this is making some magical assumptions . Some degree of patience is unavoidable.
Leaving aside friction, the tiniest amount of torque could eventually bring the body up to the desired rate of rotation; the question is how long we are willing to wait. The units for torque are force times distance, but the "distance" can't be treated as in other situations; torque is not the same as energy or work, even though the units are the same. Torque is better thought of as moment of inertia (kg-m\(^2\)) times angular acceleration (rad/s\(^2\)). Let \(a\) be the angular acceleration. then \[ \omega = at \] is the rate of rotation at time \(t\), and the moment of inertia was noted above.
The value for \(\omega\) we wish to reach, at full speed, is \[ \omega = 2\pi R = 2\pi \left({S\over 2\pi r}\right) = {S\over r}.\] From \(\omega = at\), it follows that the angular acceleration must be \[ a = {S\over rt}. \] The torque required to reach speed \(S\) in \(t\) seconds for a mass \(M\) of radius \(r\) is then \[ \tau = I a = {1\over 2} M r^2\cdot {S\over rt} = {MSr\over 2t}. \] Since \(S\) is at most 16 m/s, this becomes \[ \tau = 8Mr/t. \] If we know how much torque is available, this tells us how long it will take to reach full speed; or, if we how how long we want to wait, it tells us how much torque is necessary.
When run at variable speed, the torque developed by an induction motor depends on the RPM, and there is a particular RPM at which the motor develops peak torque. This is the torque normally quoted in the motor's documentation. We're going to gear down from this optimal RPM, and gearing down (as opposed to slowing down the motor itself) by a given factor increases the torque by the same factor. That is, if the output RPMs go from 1750 to 175, then the torque rises by a factor of \(1750/175 = 10\).
Since power is torque times angular velocity, a perfectly efficient motor will produce torque at a given RPM equal to power divided by angular rate of rotation. That is, \[ \tau = {60 P\over 2\pi \cdot {\rm rpm}}. \] Since 1 hp = 746 W, this works out to a torque of 4.07 at 1750 RPM. Call it 4 N-m. In practice, motors develop this torque at somewhat less than their nameplate RPM. To be conservative say that the maximum torque, also known as "breakdown torque," occurs at 1500 RPM or 25 rev/s.
It was noted above that the maximum rate of rotation will be when \(S=16\) m/s, so that \[ R = {8\over \pi r}. \] This is the desired rate of rotation after gearing down. The gear ratio will be \(25/R\), and the torque will increase by this factor, so that the torque produced by a 1 hp motor at this gear ratio will be \[ \tau = 4\times {25\over R} = {25\pi r\over 2}. \] Equate this torque to the torque/time relationship given above, \[ {8Mr\over t} = {25\pi r\over 2}, \] to obtain \[ t = {16 M\over 25\pi},\] which is independent of the radius. As a test case, assume that \(M = 150\) kg, for an object that is large and heavy, but not outrageous. Then \[ t = {16\cdot 150\over 25\pi} \approx 31{\rm\ sec}, \] which is depressing. This is much more than the "magical" value given earlier. Even if a 5 hp motor is used, it would still take more than 6 seconds.
There's another problem. In theory, and with patience, a small motor could bring the object up to speed, but the rotation has to be able to start from zero. If the power passes through a variable transmission, then the gear ratio can start out as high as necessary to begin the rotation, then change gradually as the object speeds up, but what if the gear ratio is fixed? How much torque is needed to get things moving? Ideally, we should take into account the "breakaway" torque, which is the torque needed to overcome the initial stickiness of the assembly, but I have no idea how to determine that. Worst case, I suppose you could give the object a start by turning it by hand. Neglecting the issue of breakway torque, will the motor have enough torque to get things going? What is the "standstill" or "starting" torque?
There are equations for the behavior of induction motors, which I don't really want to unravel, but the starting torque is about 1/3 or 1/4 of maximum torque. So, if the torque of a 1 hp motor is 4 N-m, then the starting torque is about 1.2 N-m; as the motor spins up, the available torque increases. By the same reasoning as above, after gearing down so as to reach \(S=16\) m/s, the starting torque of a 1 hp motor, applied at the shaft, will be \[ \tau = 1.2\times {25\over R} = 3.75\ \pi r. \] This observation has a couple of implications. First, because a motor's torque is relatively flat until it approaches the optimal RPM, the estimate of time given above is too optimistic by a factor of at least 2. Burning up the motor while waiting for the RPMs to increase is a real risk. Second, as the rotation starts, there is not a lot of torque. If \(r=0.5\), then the initial torque is \(\tau \approx 5.89\) N-m, which isn't very impressive.
The bottom line: No realistic supply of power will be able to spin up an object weighing more than about 200 lbs in a reasonable amount of time. This still allows turning some big pieces. At 40 lbs/ft\(^3\), a maple disc six feet in diameter and two inches thick weighs 188 lbs, and a hemisphere two inches thick and four feet in diameter weighs 168 lbs.
There are two issues to be addressed: providing enough power and torque, and allowing for a widely variable gear ratio. Above, a standard induction motor was assumed. Maybe there are other possibilities.
For example, certain treadmills have the necessary parts. In fact, the lathe made by Ray Makes, which is similar to what I plan, uses a Shimpo variable speed cone drive transmission which he pulled from a "medical grade" treadmill. That's a great solution if you have one of these transmissions, but I don't.
A DC motor has a better torque curve than an induction motor for starting heavy loads, but those are harder to pick up used, and it doesn't solve the larger problem of providing enough power. Power is power, and even under magical assumptions, there's not enough to get a large object up to speed fast enough. Even if I happen on a large DC motor at a bargain price, there's still the problem of providing all that DC current.
Stepper motors also produce maximum torque at low RPMs, but the largest stepper I could find, on a NEMA 42 frame, has a maximum torque of 28 N-m, which sounds pretty good, but the torque falls off to about half that at 400 RPM. Thinking in terms of HP, this is only about 0.8 hp (\(14\cdot 2\cdot\pi\cdot 400/(60\cdot 746)\)), so this option is out on the basis of "power is power."
A moderately crazy possibility is using the hydraulic pump on my tractor. This would provide plenty of power and torque, and it's possible to set up hydraulic gear motors to make it easy to vary the output RPM. This is tempting, but I'd rather not deal with it. Maybe...someday...
We're down to using a standard induction motor, and now the question is how to gear it to allow for a range of RPMs from 10 to 1000. A VFD is obviously the first step, but these only allow running the motor from about 50% to 150% of rated RPMs. With some motors under some circumstances, it may be possible to go outside this range, but 50% to 150% is conservative. The big risk, at low RPMs, is overheating. For a four-pole motor, the most common type, the VFD thus provides a range from about 800 to 2400 RPM. We need some serious gear reduction. In rough numbers, the range needed is from 100-1 to 2-1.
I will continue to noodle this while building the framework, but some kind of belt-and-pulley system seems easiest, perhaps with a friction clutch (like on a lawn tractor) to allow a low range for maximum torque starting out, then a high range.
A 100-1 ratio either means using huge pulleys or several smaller pulleys. The smallest pulley one could reasonably use is roughly 1.75 inches in diameter. If there are two stages, then the second pulley needs to have a diameter as large as 175 inches or more than 14 feet! If there are three stages, then we need two ratios of 10-1, and even a 17.5 inch pulley is too big. With four stages, each ratio needs to be only about 4.6-1 (\(\sqrt[3]{100}\approx 4.64\)). That's doable, even if it's a pain to have so many stages.