How does drill variable speed work?

[Chris Smith]

No.

Almost but not 100%.

Friction on earth, etc as described from these posts will not allow it to be the *same*, using two sets of paramaters with the same amount of power.

*"If"* the HP of the motor does exceed both load possibilities, then it will achieve its peak RPM "regardless" of ether "small" load,.... but it will slow down and not achieve maximum speed if the load increase [adding more mass] exceeds its HP rating.

[rshayes]

It sounds like the bearing friction is rather high. If that PCB drill has sleeve bearings, then the bearing friction for the heavier disk will probably limit the maximum speed to a lower value.

The disks may be different sizes as well as weight. When you try to estimate angular accelleration, the information that you need is the moment of inertia rather than the mass. The moment of inertia is proportional to the weight of a disk, and is also proportional to the square of the diameter. So shape may be a more significant factor than weight.

The total moment of inertia also includes the inertia of the motor armature.

The total moment of inertia and the total drag forces define a mechanical time constant. This is similar to a capacitor discharging through a resistor. If the bearing friction and the moment of inertia were both proportional to the weight of the disk, the time constants would be the same, and they would coast down to a stop in the same time. Since this isn't the case, it would appear that the heavier disk is also larger than the lighter disk.

The time required to reach the same speed will be longer for the disk with the longer time constant. Whether this happens in 10 seconds or not depends on the degree of approach to the final speed and the mechanical time constant.

The final speed is the point where the motor torque balances the frictional drag of the load. The torque of a permanent magnet motor is closely proportional to the armature current. It will be maximum at zero speed. At this point, the armature current is the applied voltage divided by the armature resistance. When the motor turns, the armature generates a voltage (often called "back EMF") which opposes the applied voltage. The armature current is now the difference between the applied voltage and the back EMF divided by the armature resistance. This makes the motor torque decrease as the motor speed increases. The torque avaliable for accelleration is the motor torque minus the drag forces.

If the drag forces are higher, the final speed will be lower, since the additional torque will require more armature current, which will increase the voltage drop in the armature resistance. This increase in voltage drop results in a decrease in the speed where the applied voltage, armature drop, and back EMF all sum to zero. (This is also the point where the torque is equal to the drag.)

Since the bearing friction is higher for the heavier disk, the final speed will be lower. Note that ball bearings might result in a different result if their friction was less than proportional to the weight of the heavier disk.

To a first approximation, the motor and disk will act like an RC circuit with a time constant determined by the ratio of the moment of inertia to the drag forces and with the final velocity determined by the motor characteristics and the drag forces.

Coasting down is simpler, since it only involves the initial velocity and the time constant.

<small>[ February 23, 2006, 06:57 PM: Message edited by: stephen ]</small>

[Dean Huster]

"Dean

You obviously have a circus near by you?"

Huh??????

I think I've figured out how things went awry ....

[MicroRem]

initial question: "How does a variable speed drill work?" Without doing all the math, and with my apologies to Einstein, I was able to determine...

answer: they work great!

let's bury this thread and move on... maybe we could get a thread going about practical electronics.

best regards to all

Tom

[MrAl]

Hi again,

Tom:
Oh gee, and take all the fun out of this
argument?

Take care,
Al

[labview1958]

I am using a standard pcb drill. The heavier disc weighs about 700g and the lighter disc weighs 50g. I am using 6V for both to reach maximum speed. I do not have a tachometer to see the rpm. A casual glance appears to show similiar maximum speed. If not, should I increase the voltage of the heavier disc to 7V or higher to reach the maximum speed for 6V of the lighter disc?

[Chris Smith]

No, I don’t think its made for all that?
Your rpms are similar, less but similar, and why do you need that exact speed. The larger disk has a cutting edge surface that is much faster because of it diameter, so why?

If you speed it up you risk it blowing up.

[MrAl]

Hello again,

That's what i was wondering too...why the
certain speed requirement?

I have several types of drills (dc and ac) and
with every single one of them with various size
grinding/cutting disks they ALL slow down when
i dig into the work, and the only way around this
is to build a speed control, which i dont feel
is really that necessary for cutting and grinding.

I built a speed control for grinders a
while back and it was very efficient because
it used a triac to control firing angle and
thus power to the grinder motor (brush type).
The main reason for the controller however was
to slow the motor down to a more managable speed,
not so much to keep speed constant...so im
wondering too now.

BTW, 50g and 700g are very different disks.
I was thinking maybe they were more the same
than that, but thanks for mentioning that now.
There is a chance the friction in the bearings
increases 14 times with the heavier disk, which
i would think should make some difference in
top speed, but it's difficult to say just how
much when we dont know the character of the
bearings, although i still think the cooling
mechanism will eat up more power than the bearings,
assuming the bearings are still in good condition.

Increasing the input voltage will work to a
point, but as Chris says, if you go too far the
wire in the motor will overheat and perhaps
burn the insulation and short out...then no
motor torque anymore.


Take care,
Al

[rshayes]

That 700 gram disk sounds like it's way too heavy for a motor and bearings designed to drill printed circuit boards. The bearings are likely to wear out very quickly.

[labview1958]

Actually I am using the PCB drill in an experiment. Both disc are similiar in size but different in mass. The disc spin vertically, "free wheeling" and are not use for any work normally associated with the PCB drill. I am using the PCB drill because it's stated rpm can go to about 10 000 rpm. I need a constant say about 3000 rpm for my experiment for both disc. I have found that 6V should be sufficent to produce a maximum 3000 rpm.

[MrAl]

Hello again,

Oh ok, so you are saying that the drill goes 3000
rpm at 6v but it can go to 10000 rpm, so that means
that it normally can take more than 6v i guess.
This means you should be able to apply more
voltage without any problem.

Here are a few equations that show the working
of the electrical equivalent of the spinning
disk...

v is voltage (or velocity)
R is resistance (or 1/friction)
C is capacitance (or Mass)
I is current (or Force)
dv/dt is rate of change of voltage
s is the Laplace variable
The electric circuit is a capacitor in parallel
with a resistor also in parallel with a current
source (constant current) where the current
level is zero at t=0 but just to the right of
zero time (t=0+) the value is I, which implies
a unit step current I at time t=0.

The equation:
v/R + C*dv/dt = I

Solved for v:
v = R*I - R*C*dv/dt

Transformed to the frequency domain:
v = R*I(1/s) - R*C*s
(note 1/s factor for the unit step forcing function)

Now applying the final value theorem that says
the final value is equal to the limit of the function
"s times the function of s"
as s approaches zero,
we get the final value of the voltage (velocity):

v(final)=lim(s-->0)[s*(R*I(1/s)-R*C*s)]

Doing the math and taking the limit, we get:
v(final)=R*I

which simply states that the final value of the
voltage (or velocity) after some long time interval
is equal to the resistance times the current.
Thus, if the resistance was 10 ohms and the
constant current was 1 amp, the final value of
the voltage would be 10 times 1 which of course
equals 10 volts.

Now as Chris had mentioned, we didnt account for
the effects of gravity on the spinning disk, so
if he would be good enough to tell us the
amount of the contribution from gravity i could
include this term in my equation. Since from
the sound of it this would be an energy dissipator,
so i would think it would enter the equation
like this:
v/R + C*dv/dt + v/Ge(x) = I

where
Ge(x)
we are calling "The earths gravitational effect
function of (as yet) an unknown variable".
The unknown (or rather unnamed) variable might
simply be the speed of rotation (velocity),
and the function would include the acceleration
of gravity somewhere (so we could adjust at
different parts of the earth or even on other
planents).

If Chris would be good enough to supply this
information i would be able to convert it into
a function so we could estimate its effect on
the spinning system. We could then get a good
idea how significant it is and either dismiss it
or make sure to include it in any future calculations.


Take care,
Al

[Chris Smith]

Im sure the gravity equation in a mere mass of 700 grams with these rpms should be less than .00000000001% over all resistance directly caused by earths gravity , not including bearing friction or the weight of the air surrounding it causing resistance.

When your mass gets up there with the moon at 7.36 × 1022 kilograms you can use the equation of time instead of resistance to figure out the slowing rates of objects like the earth and rate it in the form of eons instead of mere millennia.

But earths gravity accounts for almost all of the friction in the bearing and thus accounts for the speed difference of the two masses at two speeds, with the same power factor.

As to spinning a 700 gram object at 10,000 rpms, Make sure you are behind Plexiglas or bullet proof glass at the very least.

700 grams at 10,000 Rpms when it comes apart will knock a hole through a wall.

I know, I enjoyed doing that one time about 30 years ago with a grinder disk made of stone.

That’s why they list the maximum speed on the side of the disk these days.

[MrAl]

Hi Chris,


Oh ok, so your saying that the effect of gravity
on this spinning disk (700 grams 3000rpm) is so
small it can be completely ignored for this
problem then?

I always wear safety glasses, for what it's worth.
I've seen FAR too many problems that could come
up from not wearing them, including but not
limited to:
1. Exploding 100 amp transistors
2. Exploding 300 volt electrolytic capacitors
(sound like a cannon going off)
3. Circuit board (PCB) traces vaporizing and
flying off into space (possibly toward the eyes)
4. Resistors exploding and spewing hot carbon
fragments off into the air in all directions

The transistors usually spew hot plastic and
molten metal and hot silicon out in various
directions. What a mess, and dangerous.

As you can see from the above i've seen my share
of problems that can and do and will occur so
i always wear eye protection. I recommend this
to everyone else too...i'd hate to hear of any
accidents occuring while trying some of the stuff
we talk about in this forum.


Take care,
Al

[Chris Smith]

Al,......That portion of the gravity equation can be put aside for the argument sake of what he is trying to achieve, yes. It is too small to consider.

But remember the bearing friction is also a product of mass caused by gravity and the spinning and the direct downward force types of movements, hence all the basic friction here is a product of gravity.

Our motion of this earth makes every “object of mass” increase by the speed of our travels through the cosmos, including our orbital spin. The 700 grams in question is a direct product of our gravity and our velocity through space. If the earth were to slow down by 10% over all, the experiment’s bottom line would alter proportionally with that change in over all mass.

If you were to do this same experiment on the moon, the mass would be 1/7th and all of the frictional effects mentioned above would be reduced accordingly.

BUT, if you spin that mass at 10,000 Rpms, DUCK!

Glasses will only save your eyes, its the rest of you that makes a large target.

[MrAl]

Hi Chris,

If you are admitting that gravity has little effect
other than on the bearings friction then why
the argument previously?

Take care,
Al