Re: motor sizing vertical slide EDM

tomp-tag <tomp-tag@...> wrote:
Mariss,
Your's are alway more complicated ;-)
(math & physik challenged here )

>Message 7
> From: "Mariss Freimanis" mariss92705@...
> Date: Thu May 11, 2006 8:54am(PDT)
>Subject: Re: motor sizing vertical slide EDM
>
>TomP,
>

i f a = v/t then the 40 mS is the t
and i was just looking at the linear acc of the driven axis
never looked much at the motor ( rotary) itself

>Not quite sure how to reconcile "15Krad/sec^2" with "40mS response
>time" and "8X acceleration that I stated".:-)
>

arrg radian ( tomp looks up radian... think it's like 56 degrees or
so yeh?)
yeh 57.295, angle subtended by circumference equal to the radius....
so 15000rad/s^2
is near 262 revs/sec^2
----------------------------------------------
There are 2-pi (6.28) radians per cycle, revolution, etc. Radians are
a unit of measure just like inches, kg or volts. 15,000 / 6.28 is
2388, not 262. 15,000rad/sec^2 means a shaft will be turing 143,312
RPM after one second (15,000/6.28 = 2388 revs/sec.).
-----------------------------------------------
meaning, a blurr, and a real kick when it accelerates... ok

>I use 15Krad/sec^2 as very respectable acceleration goal on a
>servomotor. This reflects a very good motor torque to inertia ratio.

ok, so it's a cap

>This value is motor-bound in that all mechanism acceleration rates
>will be limited to or below this value.
>

2"/sec = 120"/min ok
I'm thinking I can direct couple now
looking at R+W http://www.rw-couplings.com/

>You said the goal is 2"/sec on a 5TPI screw (600 RPM) reduced from
>the motor's 3,000 RPM via 5:1 reduction.

no more 5:1 see above, just truncate the top rpm and take the slow
speed response and full torque at lower velocity range ( whatever's
left over after getting the best slow speed response )
---------------------------------------
Put that 5:1 reduction back in.:-) Without it you get only 20% of the
power (power = torque * RPM) the motor can give.

Second, you are moving a large mass. Without reduction you will have
a terrible inertial load to motor miss-match. A 5:1 reduction
decreases the reflected moment of inertia 25-fold (5^2).
-------------------------------------------

>

Yep, the acc time would be 1/2 my design limit ( thats good )
(I hoped for 40 and you say 20, kewl )

>Using that accel rate, the motor will take 21mS to reach 3,000 RPM.
>The linear velocity will go from 0 to 2"/sec in the same period of
>time if I understand your mechanism correctly.
>

this math is all new to me..
v = 120"/min
t = 21ms = .021S
a = v / t
= 120 / .02
no, in seconds.. ( inches per seconds and acc time in seconds.. )
= 2 / .021
= 47.61904761904762 ???
-------------------------------------------
No. 2" per second / .021 seconds = 95.2" / sec^2 (check your math)
-------------------------------------------
anyway, the same magnitude area where you were..

>Solving for "a" (a = v / t) gives 95.5"/sec^2 or 0.249 G (G =
>384"/sec^2).
>

but it seems to be .126G ( 47.5"/S^2 / 384 )
----------------------------------------
95.2" sec^2 / 384 = 0.248 G
----------------------------------------
huh? what happened to the 9G slam we had yesterday??

----------------------------------------
Wasn't 9G, was 5G, was based on your "3,500 lbs" number 3,500 / 700 =
5G
----------------------------------------

>The power needed to accelerate a 700lb mass at 0.25G to velocity of
>2"/sec is only 39.5 Watts, the energy invested is only 0.83 Joules.

0.83 joules! ???
1 joule = 1 watt second
= 100V at 1 amp for 10mS
--------------------------------------
Yep, and it's all true too.:-)

1) Watts = IPM * Lbs / 531
2) 0.25G acceleration on a 700 lb mass requires 175 lbs (700 * 0.25G)
3) Plug into (1) and get 120 IPM * 175 lbs / 531 = 39.5 Watts
4) 1G acceleration is 23,040 IPM / sec^2, 0.25G is 5760 IPM / sec^2
5) time to 120 IPM is 120 / 5760 = 0.0208 sec.
6) 1 Joule = 1 Watt-second. 39.5 Watts for 0.021 sec = 0.823 Joules.

I did make a mistake. Power increases linearly with speed. It's zero
at 0 IPM, 19.75 Watts at 60 IPM and 39.5 Watts at 120 IPM. The
average is obviously 19.75 Watts.

In (6) it should have been 19.75 Watts for 0.021 sec = 0.411 Joules
instead.
It's like leaving out the "2" in "energy = mass * velocity^2 / 2).

The amount of energy invested as kinetic energy is minuscule at
2"/sec. It is less than 1 Watt for half a second; were it dissipated
in a 1/2W resistor you held between thumb and forefinger, you would
probably not even feel it. It seems counterintuitive but it's true.

On the otherhand, if the same 700 lb mass were moving at say 600 MPH
airliner speed, the energy invested would be 11.46 million Joules.
That amount of energy would power a CNC machine (1kW?) for over 3
hours. Shows the power of the velocity squared function.:-)

Mariss
---------------------------------------

ssome magnitude problems here (likely on my end )
but the jist is
the motor I suggested before is WAY outta line, right?

>
>Mariss

thanks
TomP