I would like to determine the therorectcal acceleation of a Gantry
with the following specs
A Motor drives a gearbox attached to the gearbox is a pulley which
drives a toothed gearbelt.  The belt is attached to the gantry how
fast does it accelerate?
Motor = 1 HP
Gearbox Ratio 11:1
Pulley Dia = 4"
Gantry Weight = 660 lbs

Accell  = ?  in/S^2
Show all Steps and equations in Procees
Solve in 2 ways 
1) Using English Units
2) Using Metric Units

Gearbox effiency = 100 % , Full torque is availbe through out the range
Also 
What is the force on the belt pulling the gantry in lbs ?
What is the mass of the gantry in slugs ?  in KG's ?

This is actually a very simple question.

For a motor, power = rpm x torque
Power = force x velocity
Power = (force x distance) / time

Therefore, if you have a motor of a given POWER and a gearbox of 100%
efficiency, then your gearbox is irrelevant and can be ignored.  If
you have a motor of a given RPM or of a given TORQUE, or if your
gearbox has friction losses, then you need to take the gearbox into
consideration.  So, throw it out - we don't need it.

Pulley diameter is also irrelevant.  Quite frankly, we don't care. 
That is to say -- I am assuming that since the pulley mass was not
given, we are not concerned with pulley inertia.  So throw that out
too.  (If pulley mass is given, then it becomes a slightly more
complex problem, but if is it massless, then this problem is super
easy)

So here's what we know, or rather, what we care to know:

motor = 1 hp
gantry weight = 660 lbs

Before we go any further, let's do some quick conversions and make
things easy from here on out.

1 HP = 746 Watts (W) = 550 ft*lb/sec
660 lbs(mass) = 299.4 kilograms = 20.5 slugs

(you can get these conversions yourself by going to the google main
page and typing "660 lbs to kg" or "660 lbs to slugs" and so on and so
forth)

On earth (which is were I am assuming this problem is based), g = 9.81
m/sec^2 = 32.2 ft/sec^2

Therefore, 299.4 kg exerts a force of 2937 Newtons (N)
20.5 slugs exerts a force of 660 lbs

So, now the tricky part.  We want to determine the acceleration. 
Since we have constant power output by the motor, we can figure out
how long it takes for the ganrty to travel a given distance and then
use the standard relationships between acceleration, velocity, and
time to determine acceleration.

For the metric calculations, let's make this distance 1 meter
For the English unit calculations, we'll make it 1 foot.

Now get this - 1 hp can raise 550lb by one foot in one second.

Therefore, it would take 660/550 seconds to raise 660 lbs by one foot.

660/550 = 1.2 seconds.

For a more mathematical approach, consider this:

power = (force * distance) / time
550 ft*lb/sec = 660 lb * 1 ft/ time
time = 1.2 seconds

So, let's use this equation of motion:

deltax = Vot + 1/2 at^2

where deltax = distance travelled = 1 ft
Vo = initial velocity (zero in this case)
a = acceleration
t = time = 1.2 seconds

solve for a...

1 = 0 + (1/2)*a*1.44
a = 1.39 ft/sec^2

You can do the same thing for the metric calculations.

As far as force in the belt:

F = m*a = mass * acceleration

You have two components to the acceleration:  1) a downward
acceleration due to gravity and 2) an upward acceleration due to the
motor.  YOU MUST ADD THESE TOGETHER!

Therefore, 
F = 20.5 slugs * (32.2 ft/sec^2 + 1.4 ft/sec^2)
F = 20.5 slugs * (33.6 ft/sec^2)
F = 705.2 lbs

Again, you can do the metric calculations yourself.

Unfortunately, I am not a google answerer, so there goes a potential
$20, but I hope it helped you out.

-touf

Ooops = didn't realize you wanted the answer in in/sec^2

In that case, just convert ft to inches, and you get 1.39 ft/sec^2 = 16.7 in/sec^2.

Cheers!

One quick disclaimer --- the way I did the problem assumes we are
working against gravity...the gantry moves in the direction parallel
to earth's gravitional pull.

If the gantry moves perpendicular, then it's a different answer!

Touf,

Thanks for your time on the problem.  It is not exactly what I was
looking for because in this case the Gantry does move parrellel with
the Earth.  The only effect that gravity should have on the equation
is on how much force is required to overcome friction.  In this case I
am assuming the gantry runs on frictionless bearings.

The concept that the gearbox ratio and pulley dia do not matter is interesting.
I understand (at least I think I do) why you say it does not matter
but in real life it does.  Consider riding a 10 speed bike.  In first
gear You can accellerate very easily , in tenth not so easily.

The gearbox and pulley provide the final drive ratio.

My calcs go something like this:

HP = T * RPM/5252 or T = HP * 5252/RPM (Torque in ft lbs)
Since HP = 1 then T = 5252/RPM
RPM = 1700 (this is about the speed that a standard electic Motor will
develop full Torque )  I forgot to include RPM in the orgiginal specs
T = 5252/1700 = 3.089 ft lbs
But the Gearbox Ratio is 11:1 therefore torque at the Output shaft of
the gearbox is 11*3.089 = 33.984 ft lbs
The Pulley Dia is 4"  it has a raduis of 2" or 2/12 ft  = .16666 Ft
T = Force  * Dis : force = Torque/Dis = 33.984/.1666 = 203.901 lbs 
So I am assuming that I am able to apply a constant force of 203.901
lbs on this gantry.

Now  I use the Equation F = M * A or A = F/M
F = 203.901 lbs M = 660 /32.17 = 20.5 Slugs
A = 203.901/20.5  = 9.939 ft/Sec^2 = 119.26 in/sec^2

Seems kind of High to me because if Vf = Vi + A*T 
(Vi = inital Vel = 0, 
Vf =  Final Vel  = ((1700/11)* 3.1415*4)/60 = 32.36 in/sec
T = Sec = ?
Then T = Vf/A

T = 32.36/119.26 = .2713 Sec

Then id D = Vi*T+ .5*A*T^2 then 
(Vi = 0)
D = .5*119.26*.271*.271 = 4.37 inches

I shoud be able to have this gantry up to speed in 4.37 inches - 

Just seems way to Quick for me.

I will be using a Freq Drive and I know that full Torque is not
availble at 0 RPM but I am trying to come up with a model that is at
least a close to what I can expect in real life.

When I first did the problem I solved it in Real units and then
checked in metric and I got two different answers ( I found the
conversion Error).  I have fixed that problem and can solve the
problem either way and get the same answer but it just doesn't seem to
be what I am expecting.

Any help is appreciated 

Regards,

Ron H

Ron - you silly boy - leaving RPM out of the equation...

Sorry about earlier - I misunderstood the question, but now that I do, here we go:

I will be doing this first in metric units.

As usual, some conversions to get us started:

1700 RPM = 178 rad/sec
1 HP = 746 Watts = 550 ft*lb/sec
Recall that 1 Watt has units N*m/sec
4" = 0.1016 m = 0.333333 ft
660 lbs = 299.4 kg = 20.5 slugs

POWER = TORQUE * ANGULAR VELOCITY

746 N*m/sec = TORQUE * 178/sec
4.19 N*m = TORQUE

Now, we pass through a gearbox with 11:1 ratio

TORQUE AT GEARBOX = 11 * 4.19 = 46.1 N*m

Here's what you have to remember.  Torque does not accelerate
gantries;  Force does.  Torque can only provide angular acceleration,
not linear. (Technically, anyways) - So, we need to convert from
TORQUE to FORCE.

How to do this?  Simple!  Torque = Force * Distance
(where distance in this case would be the radius of your pulley)


46.1 N*m = FORCE * radius

radius = diameter/2 = 0.106/2 = .053m

46.1 N*m = FORCE * 0.053 m

FORCE = 869.8 N

Now, Newton's 2nd law says F=ma

Therefore...

869.8 N = 299.4 kg * a
2.9 m/s^2 = a

Force in your belt is the same force being used to accelerate the mass = 869.8 N


NOW, ENGLISH UNITS (not real units...this is by far the stupidest unit
system in history -- in fact, it's so stupid, not even the English use
it anymore, but that's a story for another day...!)


POWER = TORQUE * ANGULAR VELOCITY

550 ft*lb/sec = TORQUE * 178/sec
3.09 ft*lb = TORQUE

TORQUE AT GEARBOX = 11 * 3.09= 34 ft*lb

Convert from TORQUE to FORCE.

34 ft*lb = FORCE * radius

radius = diameter/2 = .333333 = .16666 ft

34 ft*lb = FORCE * 0.1667 ft

FORCE = 204 lb

F=ma

Therefore...

204 lb = 20.5 slugs * a
9.95 ft/s^2 = a = 119 in/sec^2

---------------------------------------

Time for a sanity check.

Does 2.9 m/sec^2 = 119 in/sec^2
Well, doing the math, 2.9 m/sec^2 = 114 in/sec^2
YES, they match...(you say - wait, they're off by 5! -- 
The 5 in/sec^2 can be attributed to rounding errors...I did round quite a bit.)

Now, let's see if this makes sense, from a real world point of view.


9.95 ft/sec^2 (let's call it 10) is roughly equal to:

10 (ft/sec)/sec * 1 mile/5280 ft * 3600 sec/hour = 6.8 mph/sec
So, assuming constant acceleration (yea right), we're talking 0-60 in
8.8 seconds or 0-30 in 4.4 seconds.

Let's take a typical car, let's say my Camry.

133 HP @ 5500 RPM
3000 lbs
0-60 (spec) = 11.1 seconds
0-30 (spec) = 4.1 seconds
Gear 1 ratio = 2.9

So, let's say we stay in gear 1 up to 30, and we're getting constant
torque throughout, just for kicks.  Oh, and no drag.

Ok, so my camry is 133 times as powerful as your motor, but my gearbox
is 2.9/11 times as weak, leading to a net effect of 35 times as
powerful.

My car is also 5x as heavy as your gantry, so my acceleration should
be 7x as fast as your gantry (so far).

Oh, but power on my car is at 5500 RPM, while your motor is at 1700RPM
so we take my 7x advantage and multiply by 1700/5500 = 2.17x advantage

Now, let's take the fact that my wheels are not 4" diameter; rather
like 16", so there's a 1/4 drop there, but I do have 2, so that's like
a 1/2 drop factor, making it something like a 1.1x advantage, minus
drag, etc...pretty close to a 1x factor.

Which, if you look at my spec and the figure I got for you, which are
about equal, from 0-30, then you can see this figure makes sense.

Hope that helps.

When I said "I have 2", I meant a 2" diameter driveshaft, not 2 wheels.

Touf, 

Thanks for your time and help.  We both got the same answer which
makes me feel better.
I know you said your not on google answers but if you sign up the $20 is yours.
I will do some measurements when we build it and it will be
interseting to see what the results are.
I suspect that if I am capable of accelleating at any where near  that
rate I will need to slow it down to make it smoother.
As the the English metric thing - metric may be a bit easier to
manipulate because of the decimal point thing but Engish units are
based on real life observations and work better for most of the things
that I deal with - (at least to me).

Thanks for your help

Regards,

Ron H

I'm glad I could be of help

As far as you building it, you have to remember there are going to be
a lot (and I mean a lot) of other things to deal with.

1) gearbox efficiency is probably more like 85%
2) pulley has moment of inertia
3) friction (gantry, pulley, motor, belt drive)
4) back emf in your motor
5) power supply needs to be regulated very well
...and so on and so forth

You're also assuming you're running the motor at full speed, which is
probably not recommended, as it will affect your motor life.

Add all these together, and I think you are going to be fine.  Trying
to add all these factors into the equation is going to make it rather
complicated.  Well, it's all equations, but now you're going to have a
much longer equation is all.

Probably a safe thing to do is to take that acceleration value you
got, multiply it by about a half, and expect to get that value for
your acceleration when running at full steam.  Just running a few
numbers in my head, that sounds pretty darn close, within 10% or so.

Keep the $20 - use it to build your device...say, what is this thing for anyways.

good luck.

-t