Motor Torque Question
Archiver
Posts: 46,084
I found a motor in a catalog, and I have some questions as to the
specifications that the vendor lists.
The motor says: ( Mfg specs: 41.7 / 38.8 rpm with 80.5 oz-in load ). How
does the 80.5 oz-in load
correlate to the 41.7 / 38.8 rpm numbers ? I'm planning on using it to
drive the wheels for my mower-bot.
specifications that the vendor lists.
The motor says: ( Mfg specs: 41.7 / 38.8 rpm with 80.5 oz-in load ). How
does the 80.5 oz-in load
correlate to the 41.7 / 38.8 rpm numbers ? I'm planning on using it to
drive the wheels for my mower-bot.
Comments
Motor Torque Question:
>
>The motor says: ( Mfg specs: 41.7 / 38.8 rpm with 80.5 oz-in load ). How
>does the 80.5 oz-in load
>correlate to the 41.7 / 38.8 rpm numbers ?
80.5 oz-in is a torque and 'just' means that if the radius of the
wheel on the motor shaft is R, then that wheel can exert a force of
80.5/ 16R on the ground. The 16 changes the oz unit of force into the
more common 'lb' (pound) of force. (welcome to the totally bonkers
world of US units!
A wheel with radius = 4 inches would be able to exert a force of a
bit more than 1 pound when turning at a rate of about 40 rpm.
Using the circumference of the 4 inch wheel (C = 2 pi R) tells you
that it would move forward a distance of about 80 feet each minute
or more than 1 foot per second.
sounds like you might want to gear it down a bit or choose a tire
smaller than R=4.
--
.-. .-. .-. .-. .-. .-. .-. .-. .-. .-
\ / \ / \ N / \ C / \ S / \ S / \ M / \ / \ /
`-' `-' `-' `-' `-' `-' `-' `-' `-'
Chuck Britton Education is what is left when
britton@n... you have forgotten everything
North Carolina School of Science & Math you learned in school.
(919) 286-3366 x224 Albert Einstein, 1936
load. I assume this is a gearhead motor, no? I don't think these motors have
enough power to do what you want.
Original Message
> I found a motor in a catalog, and I have some questions as to the
> specifications that the vendor lists.
>
> The motor says: ( Mfg specs: 41.7 / 38.8 rpm with 80.5 oz-in load ). How
> does the 80.5 oz-in load
> correlate to the 41.7 / 38.8 rpm numbers ? I'm planning on using it to
> drive the wheels for my mower-bot.
the RPM may speed up and vice-versa.
Another issue to consider is the maximum speed you intend for the robot.
RPM x circumference of wheels = distance per minute.
ACJacques
Bryan Smith wrote:
>
> I found a motor in a catalog, and I have some questions as to the
> specifications that the vendor lists.
>
> The motor says: ( Mfg specs: 41.7 / 38.8 rpm with 80.5 oz-in load ). How
> does the 80.5 oz-in load
> correlate to the 41.7 / 38.8 rpm numbers ? I'm planning on using it to
> drive the wheels for my mower-bot.
>
> To UNSUBSCRIBE, just send mail to:
> basicstamps-unsubscribe@yahoogroups.com
> from the same email address that you subscribed with. Text in the Subject and
Body of the message will be ignored.
>
>
> Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/
lever one foot away from the motor's shaft that the motor would rotate
this load at some given speed? One foot per second or maybe one radian
per second, or maybe some where in the range of 41.7 / 38.8RPM?
Regards,
Leroy
Rodent wrote:
>
> I think what they are saying is that the motor will run a bit slower with a
> load. I assume this is a gearhead motor, no? I don't think these motors have
> enough power to do what you want.
>
>
Original Message
>
> > I found a motor in a catalog, and I have some questions as to the
> > specifications that the vendor lists.
> >
> > The motor says: ( Mfg specs: 41.7 / 38.8 rpm with 80.5 oz-in load ). How
> > does the 80.5 oz-in load
> > correlate to the 41.7 / 38.8 rpm numbers ? I'm planning on using it to
> > drive the wheels for my mower-bot.
>
> To UNSUBSCRIBE, just send mail to:
> basicstamps-unsubscribe@yahoogroups.com
> from the same email address that you subscribed with. Text in the Subject and
Body of the message will be ignored.
>
>
> Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/
--
*******************************************************
* Leroy Hall *
* 317 Cherokee Drive *
* Loveland, Ohio - USA 45140-2404 *
*******************************************************
* Phone: (513) 697-7539 *
* Cell : (513) 300-8632 *
* Email: leroy@f... *
* Home page URL: http://home.fuse.net/leroy/ *
* Resume URL: http://home.fuse.net/leroy/resume.htm *
*******************************************************
* Leroy Hall *
* 317 Cherokee Drive *
* Loveland, Ohio - USA 45140-2404 *
*******************************************************
gravity use a spring-scale to measure pull.
The torque is relative to the length of the lever -- if you fixed an arm on
the motor shaft one inch long, it would exert 80.5 ounces of pull on a
spring scale while the motor rotated at the given speed. If the arm were
longer, the pull would be proportionally less, but the torque would remain
the same. This means smaller diameter wheels are gonna give you more power
than larger ones.
Original Message
> Does the 80.5 ounce load indicate if this weight were suspended on a
> lever one foot away from the motor's shaft that the motor would rotate
> this load at some given speed? One foot per second or maybe one radian
> per second, or maybe some where in the range of 41.7 / 38.8RPM?
> > I think what they are saying is that the motor will run a bit slower
with a
> > load. I assume this is a gearhead motor, no? I don't think these motors
have
> > enough power to do what you want.
Motor Torque Question:
>Don't throw gravity into this -- it will just confuse things. Imagine no
>gravity use a spring-scale to measure pull.
>
>The torque is relative to the length of the lever -- if you fixed an arm on
>the motor shaft one inch long, it would exert 80.5 ounces of pull on a
>spring scale while the motor rotated at the given speed. If the arm were
>longer, the pull would be proportionally less, but the torque would remain
>the same. This means smaller diameter wheels are gonna give you more power
>than larger ones.
Gravity is ok here.
An ounce of pull is the weight of a 1 ounce object.
A one inch radius wheel is a one inch lever. It could lift a 80.5 oz
load ( 5 lbs) at the rated rpm. (from this you can figure the
velocity of the lifting)
POWER is Force X Velocity as well as being Voltage X Current.
so changing lever lengths (wheel radius) will NOT change the POWER
delivered but 'just' changes the FORCE and the VELOCITY at which this
force acts.
Power and Force are sometimes colloquially used interchangeably
but I'm trying to be PC here. (Physics Correct)
--
.-. .-. .-. .-. .-. .-. .-. .-. .-. .-
\ / \ / \ N / \ C / \ S / \ S / \ M / \ / \ /
`-' `-' `-' `-' `-' `-' `-' `-' `-'
Chuck Britton Education is what is left when
britton@n... you have forgotten everything
North Carolina School of Science & Math you learned in school.
(919) 286-3366 x224 Albert Einstein, 1936
the load is suspended perpendicular to the ground. Not a good example for
explaining motor torque.
Original Message
> >Don't throw gravity into this -- it will just confuse things. Imagine no
> >gravity use a spring-scale to measure pull.
> >
> >The torque is relative to the length of the lever -- if you fixed an arm
on
> >the motor shaft one inch long, it would exert 80.5 ounces of pull on a
> >spring scale while the motor rotated at the given speed. If the arm were
> >longer, the pull would be proportionally less, but the torque would
remain
> >the same. This means smaller diameter wheels are gonna give you more
power
> >than larger ones.
>
> Gravity is ok here.
>
> An ounce of pull is the weight of a 1 ounce object.
>
> A one inch radius wheel is a one inch lever. It could lift a 80.5 oz
> load ( 5 lbs) at the rated rpm. (from this you can figure the
> velocity of the lifting)
>
> POWER is Force X Velocity as well as being Voltage X Current.
>
> so changing lever lengths (wheel radius) will NOT change the POWER
> delivered but 'just' changes the FORCE and the VELOCITY at which this
> force acts.
>
> Power and Force are sometimes colloquially used interchangeably
> but I'm trying to be PC here. (Physics Correct)
I | \
| \
____\_____
N
OK, class: The chart above is typical Torque (and current demand) of
a typical DC motor, versus speed (N), at constant voltage.
A responsible motor seller will tell us about TWO points on the line.
Often we know "free rpm" at a given voltage; the motor hooked to
battery with no output load (output torque is zero).
Grab the shaft, speed goes down, torque and current increase. Stop
the shaft ("stall"), you'll have max current flow and torque. Seller
should give "stall torque" and "stall current". Buyer needs to size
his speed control for the max stall torque current, if stalling is
possible (mower stuck in a ditch), or use a speed control with some
kind of current limiter (e.g., Allegro 3952).
Typically, a DC motor has best efficiency, and is sized to run,
around Nmax/2.
The rating given below isn't clear- is this max speed and stall
torque? Or actually a point on the line?
Calibration point: Typical cordless screwdrivers will turn max of
around 180 rpm (free speed), and give about 40 in-oz stall torque.
These gearmotors are often used by folk who motorize GI Joe vehicles
for R/C use (yup...).
You need to decide how fast your Bot needs to go, and select an
appropriate wheel diameter, which yields the needed output speed...
which should be about half the free speed. The torque numbers below
seem to be in the ballpark.
Check jameco.com for a nice selection of gearmotors (search for dc
motors)... or:
http://jameco.com/Jameco/Products/ProdCT/p078.pdf, bottom of page.
patmat
--- In basicstamps@y..., Bryan Smith <bsmith@w...> wrote:
> I found a motor in a catalog, and I have some questions as to the
> specifications that the vendor lists.
>
> The motor says: ( Mfg specs: 41.7 / 38.8 rpm with 80.5 oz-in
load ). How
> does the 80.5 oz-in load
> correlate to the 41.7 / 38.8 rpm numbers ? I'm planning on using
it to
> drive the wheels for my mower-bot.
>
Original Message
> From: Bryan Smith [noparse]/noparse]SMTP:[url=http://forums.parallaxinc.com/group/basicstamps/post?postID=dEPtyYr9x3VlvJqBQZNNMQ_VeY6EVN15RJ4W_yjveH5UWO3_TUSL4Exn0cCHewTUik2eXbHFmL8]bsmith@w...[/url
> Sent: Tuesday, June 19, 2001 3:46 PM
> To: basicstamps@yahoogroups.com
> Subject: [noparse][[/noparse]basicstamps] Motor Torque Question
>
> I found a motor in a catalog, and I have some questions as to the
> specifications that the vendor lists.
>
> The motor says: ( Mfg specs: 41.7 / 38.8 rpm with 80.5 oz-in load ). How
> does the 80.5 oz-in load
> correlate to the 41.7 / 38.8 rpm numbers ? I'm planning on using it to
> drive the wheels for my mower-bot.
>
>
>
> To UNSUBSCRIBE, just send mail to:
> basicstamps-unsubscribe@yahoogroups.com
> from the same email address that you subscribed with. Text in the Subject
> and Body of the message will be ignored.
>
>
> Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/
>
Why two numbers are used? Is the second one the RPM's
with load?
[/font]
patmat
--- In basicstamps@y..., parkan197@a... wrote:
>
>
> > 41.7 / 38.8RPM <
Why two numbers are used? Is the second one
the RPM's
> > with load?