[Semus]

I use a Bosch 1617 in my CNC router machine, and I'm currently using HSMXpresss + HSMAdvisor for all of my cutting needs.

I was wondering if anyone knew what the max torque was for this router? I don't know if it's super critical, but HSMAdvisor has the ability to add a specific machine, and it's asking for things like max and min RPM's, max HP and max torque. Again, I don't know if max torque is critical, but I'd just like to know if anyone had any idea what it could be. I haven't found any info online about the torque with this router.

Is the max torque entry critical to getting accurate speeds/feeds using HSMAdvisor? I was able to enter all the other data, so I was just wondering if anyone had the torque info, just to be safe.

[alan_3301]

2.25 HP @ 25,000 RPMS = 5.64 inch pounds

5.64 * (RPM / 25000) = max theoretical torque at specified rpm, in reality will be less.

1.8 inch pounds @ 8000 rpms

If my math is right. (Barely finished high school )

[mkbryant]

Semus, is this the Fixed-speed 25,000 rpm version of the 1617, or the common one I see everybody using, the Variable-speed 1617EVS? I think the EVS has a min speed of 8000 rpm, but if you use a SuperPID you could theoretically get it down to 5k rpm.

Sorry to disagree with you alan_3301 (well, partly), but the Power-Torque-Angular velocity equation is:

H = T * w,
where:
H = power in base unit (W for SI, ft*lbf/sec for USCS)
T = torque in base unit (N*m or ft*lbf)
w =angular velocity (lowercase omega, rad/s).

So solving for T, you get:

T = H/w.

Torque to angular velocity is an inverse relationship. This means that for fixed horsepower, which would be the 2.25 hp maximum, higher w would generate less torque, and lower w would generate higher torque. If your base units are RPM, hp and you want torque in*lbf, with conversion factors it would be:

T_inlbf = H_max * 33000 * 12 / (2 * pi * RPM)

• 5000 rpm >> T = 28.361 in*lbf
• 8000 rpm >> T = 17.726 in*lbf
• 10000 rpm >> T = 14.18 in*lbf
• 15000 rpm >> T = 9.45 in*lbf
• 20000 rpm >> T = 7.09 in*lbf
• 25000 rpm >> T = 5.67 in*lbf

[alan_3301]

Again, I am no mathematician or engineer, but when at lower rpms, the power will be lower, so will the torque.
It isn't 2.25 hp at all rpms. It probably isn't 2.25 horsepower at any rpms, but if that is what the marketing department gave it, then that is what I used for the conversion.

[mkbryant]

Again, I am no mathematician or engineer, but when at lower rpms, the power will be lower, so will the torque.
It isn't 2.25 hp at all rpms. It probably isn't 2.25 horsepower at any rpms, but if that is what the marketing department gave it, then that is what I used for the conversion.

The problem is that you're thinking about power incorrectly. Power is defined (by Wikipedia) as:

...the rate of doing work. It is equivalent to an amount of energy consumed per unit time. In the MKS system, the unit of power is the joule per second (J/s), known as the watt.

"Work" is done by a constant force on a point that moves a displacement in the direction of the force. In rotation, work, W, is the product of torque, T (force x radius or moment arm) and swept angle, φ. Torque can be considered the rotational analog to the linear "force". Whereas a force causes a mass to accelerate in the direction of the force, a torque causes a mass to accelerate radially in the direction of the torque. Also, when a force has accelerated a mass to a desired velocity, as long as there is no friction or counter forces, the force stops acting on the mass and the mass continues moving at that velocity until another force acts on it, in the same direction or any other. Torque is the same way. When a motor spins up a bit, a torque acts on the bit until the bit reaches a desired rotational speed (RPM). Assuming the bearings are frictionless, the bit will spin at the desired rpm until and unless otherwise acted upon.

In the real world, we know there is no such thing as frictionless bearings. Therefore, once reached the desired speed, the torque will decrease but must still act upon the bit to counteract the losses due to friction, Ff. This torque should be considerably less than that required to accelerate the bit to the desired speed. The motor is doing constant work (assuming friction is constant) and therefore the power required is constant at this stage. The interesting thing here is that, due to the work equation, W = T x φ, being dependant only upon torque and angle, it can be derived that the work is the same whether the bit is rotating at 5rpm or 50000rpm. This assumes, of course, that there is no viscous friction that changes with speed. Therefore, we can state that the work required for one rotation of the bit is independent of speed, and therefore the same for all speeds.

However, because power, P, is P = W / t, where t is time in seconds, we can combine power and work equations: P = W / t = T x φ / t = T x (φ / t). The part, φ / t, can be simplified to angular velocity, ω which has the unit radians per second. This makes the power equation for rotation P = T x ω. Therefore, because we proved earlier that the torque required to keep a bit spinning is the same at any speed, the only variable in this equation is angular velocity. It is a linear relationship, so with the same torque, lower velocity requires lower power, and higher velocity requires higher power.



These are completely made up numbers for illustration purposes only, but let's assume it takes 0.25hp to keep a bit free-spinning at 5k rpm, and 1.5hp to keep a bit spinning at 25k rpm. This is the minimum amount of power required to maintain velocity, and we're not even cutting anything yet. So if you have a maximum of 2.25hp available, that means at 5k you have 2hp that can go into cutting something, but at 25k you only have 0.75hp that is available to cut.

If you reverse the power equation, solving for Torque, T, you get T = P / ω, so for a constant ω, if you have less available power, the torque output is less. This was illustrated in my first post. But to take the example further, let's say you are using a 1/4" bit, so you have a radius of 1/8" (0.125"). Torque is F x d (or F x radius), so you can solve the power equation for Force at the cutting edge, Fc = P / (ω x r). Force and power have a linearly proportional relationship. With more power available for cutting at 5k rpm (2hp), you can generate a higher force at the cutting edge. This means you can cut tougher materials and make deeper cuts without running out of power, because these tougher materials and deeper cuts require more force than weaker materials and shallower cuts.

All this falls in line with practical knowledge. You don't cut steel (typically, not talking about High-speed machining here) at high rpms, but you can cut plastic or wood at very high rpms.

The bottom line here is that the maximum power value associated with a product needs to be thought of as available power, not as constant, always on power. Power is a quantity that is drawn as needed. In this case, power used is electrical energy that is converted to heat (resistance in the motor and wires) and kinetic energy (motor spinning). You cannot draw more power than you are using. You must have a load (spinning bit, cutting metal) or you might have a short or some sort-of inherent resistance in the system, such as a connector, length of wire, resistor, etc. Otherwise, the system is not drawing power because it does not need to do work.

[keebler303]

I'm going to agree with Alan here. A router has a peak power of XX HP, it is NOT constant over the RPM range and will be very low at low speeds. It is not possible to get 2.25HP of POWER out of a router at 5k RPM. Even if you have more torque at lower speed, you have much less power so you aren't going to get as much work done (metal removal rate).

A motor at max torque and zero speed does zero work at peak current. 0% efficiency (100% heat!). With low torque motors like the "universal" motors used in routers, you need to keep the RPMs up to keep the power and efficiency at optimal levels.

[mkbryant]

Alright, so due to the fact that I am, admittedly, not an electrical engineer, I was a bit confused about how these hand-held routers actually control and vary the speed. But I am a Mechanical Engineer by education, trade and passion, and I have had some electrical engineering classes and training. It's not in my nature to simply take someone at their word, even if they are my elder, boss, manager, or other in a position of experience and knowledge. They may be right, but I have to fully understand why what they're telling me is right. It's not because I think they're wrong, it's because I'm trying to understand why they're right. Anytime someone tells me the reason something is right because they said so, or because they just know, I set out to prove them wrong. Many times what happens is I discover that they were, in many cases, right, but I also learn why they were right. Many times, they don't even know why, they just know.

In this case, I am building a router and I plan on using a single-speed Bosch 1617 (25k rpm) in conjunction with a SuperPID Router Speed Controller. This topic is important for me because I would like to learn as much as possible about router power, cutting feeds and speeds and in particular, the actual router I am using.

I have been reading through much of RomanLini's SuperPID thread (http://www.cnczone.com/forums/spindl...ing-forum.html) and in that he discusses many times about power and torque available at low RPMs vs high RPMs. It appears that with all of these home/commercial wood routers with variable-speed controls, the way most speed controllers work is they simply control the amount of current supplied to the router motor. They supply only the amount of current (and thus electrical power) needed to spin up spindle under no load to a fixed rpm. If the controller is not smart enough to vary the current when the motor senses additional load, then the spindle will likely stall at very low rpms due to the low supplied current (power).

However, advanced controllers like the SuperPID use an optical tachometer to measure actual rotational speed, and will, when set at a specific rotational speed, vary the amount of current supplied to the motor. This means that you get increased power and thus higher torque at low rpms. RomanLini did mention that the SuperPID will not be able to supply full power to the motor at low speeds, and I still do not understand why this is. I understand there are inefficiencies inherent in electro-mechanical devices. I'm sure this has something to do with the shape and interaction between the rotor and stator, but, as I mentioned before I am not an electrical engineer, this simply escapes me. If anyone could explain exactly why these inefficiencies exist, I would love to hear/read about this.

I tried the googles to search for anything about "universal" motors and torque vs speed. These universal motors typically have brushes, as opposed to the brushless variety. I found two different sites that had decent looking graphs and reasonable explanations.

Motor Speed-Torque curve reference graph :: Groschopp tech. resources
Attachment 262510

https://www.jmag-international.com/c...teristics.html
Attachment 262512

These graphs do show drastically higher torque available at lower rpms but these appear to be theoretical and do not reference specific speed control methods. Once again, I still do not understand why you couldn't simply supply the maximum current to the router under load and therefore be able to use the full power available. I'll say it again, I do not understand this. Is it heat? Is it back emf? Is it something in the way the actual motor components are manufactured/designed?

The other thing I discovered was that induction AC motors typically have a flatter torque curve, with different inflection points (starting, pull-up, pull-out, full load). These motors are controlled by changing the frequency of the AC power supplied to them. They use variable frequency drives to control their speed.

Anyway, I'll go ahead and concede that, with these consumer wood routers with variable speed controls, the full power and max torque is not available at low rpms. But this due not to the fundamentals of physics somehow limiting these things, but in the speed control design. With better (and probably more expensive) speed controllers, such as the SuperPID, these motors will generate more torque at lower rpms due to it taking less torque to maintain speed.

[wizard]

I will try to help with this:

These graphs do show drastically higher torque available at lower rpms but these appear to be theoretical and do not reference specific speed control methods. Once again, I still do not understand why you couldn't simply supply the maximum current to the router under load and therefore be able to use the full power available. I'll say it again, I do not understand this. Is it heat? Is it back emf? Is it something in the way the actual motor components are manufactured/designed?

It has been a long long time since I took a course in college that covered this, so maybe I might screw it up but here goes. First I think you had the right idea in your earlier post. For a given torque the power out put of a spindle is directly related to it speed. This site: Power and Torque: Understanding the Relationship Between the Two, by EPI Inc. has a bit with respect to the formula. I suspect that you already have a handle on this.

So to simplify the discussion let's focus on a DC machine. In such a motor torque is directly related to motor current. The more torque being demanded the more current that flows, up to a point. The problem in the real world is that you can't pump an infinite amount of current into the windings on a DC motor, there are impedance issues and more so the heat that results from the current flow. Also in a permanent magnet motor excessive heat and current can demagnitise the magnets. Since you have this limited ability to manage current in the motor you have a maximum amount of torque that can be developed. In other words it is a very bad idea to try to pump more current into a motor than the windings are designed to handle.

Now we run into the problem of how the routers are rated. My guess is that they are rated at maximum RPM they can achieve at full load on the rotor. So at 25,000 RPM it doesn't take a lot of torque to produce the horsepower a quoted. Cut your RPMs in half and so goes your potential power output. You can increase power output at lower RPMs because you aren't free to increase current beyond the motors design limitations. At least not for long as the magic smoke will soon issue forth.

That is the nut shell explanation for a simple DC machine, that is a brushed DC motor. The idea of hitting limits though applies to other motors also. Even in an AC motor if you try to develop to much torque you will increase motor current leading to overheating and burn out. In other words you have no more torque that the motor windings will allow.

Another very significant issue that arises and is related to heat, is the ability to cool the motor windings. At low speed the built in fans often can not cool the windings effectively leading to over heating. This can limit low speed performance or require separate cooling. It isn't uncommons to see large blower motors attached to a servo motor to keep it cool through out its operating range.

As far as the more advanced router speed controllers go I suspect that they have a primary purpose of keeping the spindle RapM as constant as possible at a given RPM set point. I really don't see them deriving any more power from a motor that it is designed to deliver.

[keebler303]

That sounds logical to me. I am but an ME as well.

As far as the more advanced router speed controllers go I suspect that they have a primary purpose of keeping the spindle RapM as constant as possible at a given RPM set point. I really don't see them deriving any more power from a motor that it is designed to deliver.

The SuperPID provides some form of closed loop, a rudimetary servo drive, which should allow better performance within the limitations you have discussed. Similar to a DC motor turning at say 50 RPM compared to the same motor turning at 50 RPM under the command of a servo drive with encoder feedback. When you are open loop, you have to reduce the voltage to reduce the speed so the speed will drop when the load increases. A servo drive can increase the voltage and maintain the speed to force more current to meet the demand for torque.

[mkbryant]

So to simplify the discussion let's focus on a DC machine. In such a motor torque is directly related to motor current. The more torque being demanded the more current that flows, up to a point. The problem in the real world is that you can't pump an infinite amount of current into the windings on a DC motor, there are impedance issues and more so the heat that results from the current flow. Also in a permanent magnet motor excessive heat and current can demagnitise the magnets. Since you have this limited ability to manage current in the motor you have a maximum amount of torque that can be developed. In other words it is a very bad idea to try to pump more current into a motor than the windings are designed to handle.

...You can increase power output at lower RPMs because you aren't free to increase current beyond the motors design limitations. At least not for long as the magic smoke will soon issue forth...

Another very significant issue that arises and is related to heat, is the ability to cool the motor windings. At low speed the built in fans often can not cool the windings effectively leading to over heating. This can limit low speed performance or require separate cooling. It isn't uncommons to see large blower motors attached to a servo motor to keep it cool through out its operating range.

Wizard, thanks for the explanation. Makes things a little clearer. There seem to be so many things in the magical world of electricity that don't make sense to me. Give me a bolt, gear, spring or screw and I feel all warm and fuzzy inside but electricity makes me sad.

I hadn't thought about the overheating thing. I was looking at the router assembly diagram and realized the fan is pressed onto the shaft, and thus has a cooling capacity proportional to the shaft speed.

It seems like the main consideration here is the amount of current the windings can handle without overheating. Assuming at max speed the fan provides sufficient cooling for the windings, the maximum current is supplied to the motor. Based on this, the only reason you can't have higher torque at low rpms is due to heat load. So theoretically, not taking into account efficiency of the motor itself in converting electrical power to mechanical, if you could somehow provide sufficient cooling to the motor throughout the entire rpm range, and assuming you could dynamically supply current to the motor based on need, you could consider the motor a constant power motor, and calculate torque appropriately? Obviously, you could never supply more current than the max rated at any RPM, but you should be able to supply up to the max current at all RPMs.

I assume this is how more industrial spindles work, either with massive fins and a cooling fan, or by using liquid cooling.

I saw somebody add two small 12V blower motors to the top of their router over in the superPID thread. Looks like it worked well, but haven't seen anybody else do anything like this. Thinking of trying it, can't be that hard, right?

[ger21]

I assume this is how more industrial spindles work, either with massive fins and a cooling fan, or by using liquid cooling.

Industrial high speed router spindles (18000 to 24000 rpm) have the same problem with heat. Even with electric fan or water cooling, their low rpm range is limited because they'll get too hot.

I saw somebody add two small 12V blower motors to the top of their router over in the superPID thread. Looks like it worked well, but haven't seen anybody else do anything like this. Thinking of trying it, can't be that hard, right?

From my experience with the Super-PID, this is not necessary at all. I've run it at 5000rpm drilling holes for 30 minutes or more, and the router is still cool to the touch. Even on runs of several hours at higher rpms, it never gets hotter than just a little warm.

The Super PID seems to run the motor much more efficiently.

[mkbryant]

From my experience with the Super-PID, this is not necessary at all. I've run it at 5000rpm drilling holes for 30 minutes or more, and the router is still cool to the touch. Even on runs of several hours at higher rpms, it never gets hotter than just a little warm.

The Super PID seems to run the motor much more efficiently.

Thanks for the info, Gerry. Have you run the SuperPID at low RPMs cutting hardwoods, hard plastics or aluminum for any length of time? How did it perform temperature-wise? What size/power router are you running with it? Do you use the temperature sensor to monitor or is this not strictly necessary?

[ger21]

As I said, I've run it for 30 minutes or so at 5000 rpm. I've also run it for several hours at 12,000 rpm. I've got it on a porter cable 690.

I have a pre-release version, that doesn't have a temperature sensor. But my router barely gets warm. I don't see any reason to use the temp sensor.