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Linkage Torque/calculation

Linkage Torque/calculation

Linkage Torque/calculation

(OP)
Hi guys, im just new to this forum as a member but have been reading it for advice for a few months now, some expert advice indeed.

I am basically looking for some guidance on how to calculate the forces and torques involved in a basic linkage, which is the linkage used to curl the bucket on a back actor of an excavator. I am using it to rotate a conveyor through 180degrees. As the linear actuator (in my case an hydraulic ram) retracts it produces a moment on the arced swinging link which causes the straight link to provide a moment arm on the central large pin providing me with the rotation i need.

i would assume my ram creates a moment on the arced swinging link of 1400mm x max ram force( 53KN). At this point i dont know how to continue with the calculation to determine the maximum torque available at my main pin.

could any one tell me what i need to do to solve this problem? i have attached a jpeg of the linkage with a few dimensions i think i need

many thanks for reading and i look forward to some discussion on this

RE: Linkage Torque/calculation

(OP)
Any thoughts or suggestions at all welcome guys

RE: Linkage Torque/calculation

couldn't open your file.

how about free body diagrams ?

RE: Linkage Torque/calculation

(OP)
here is a PDF document attached which i hope can be opened rb1957 (and others).

let me know if this works guys and i appreciate all replies or ideas

thanks

   

RE: Linkage Torque/calculation

oops ... no attachment ;)

RE: Linkage Torque/calculation

(OP)
there appears to be a problem with the upload file feature on this site. is there any other way i can paste in an image or anything to this text message area?

i tried to paste a JPEG but it doesnt appear to allow that.



 

RE: Linkage Torque/calculation

RE: Linkage Torque/calculation

gmcgrory, I think if you review your statics, you will find that the moment on the large pin at any position is the product of the force applied by the actuator and the perpendicular distance pin-to-actuator. Neglecting losses.

Regards,

Mike

RE: Linkage Torque/calculation

(OP)
btrue blood, i am indeed talking about a similar linkage to the one you posted. but the one im looking at is slightly different. the curved link in mine is joined to the link on the main pin via a straight link.

what is the the dynamics of this situation? is this a difficult problem for me to ascertain what is the maximum torque i can get from my pin. is it just simply as u say the perpindicular distance from the line of force of my ram to the main pin, or do i have to take into account the other links aswell?

 

RE: Linkage Torque/calculation

having seen your pdf ... free body diagrams, two force members ... each link, for different positions ... lick o'paint

RE: Linkage Torque/calculation

(OP)
rb1957, can u explain maybe a bit further or give me an idea how to do the calculation. I done this kind of stuff in my degree but since i left uni i have not done much with regards to linkage calculations in my current work.

thanks guys  

RE: Linkage Torque/calculation

google "free body diagram", get out an old engineering mechanics text ... the force in each of your links is along a line of action between the two lug centers (at the ends of each link).  the curved arm will have some bending in it.  you'll need to rotate the geometry to understand how the loads change.  where is the load being applied ?  

RE: Linkage Torque/calculation

Like Rb says, draw your free body diagrams.  Go pin by pin, and draw the force vectors.  The curved link still transmits no moment to its fixed pin, unless I'm mistaking your diagram.  In your configuration, it is just acting to guide the motion of the actuator rod end.  Replace it in your schematic drawing as a rigid, straight rod linking the two endpoints (use dotted lines if it helps).

RE: Linkage Torque/calculation

(OP)
hi guys.. i have just read through a lengthy revision of free body diagrams and understanding it pretty well. but i am still struggling to figure out how to solve my problem.

im not sure if the diagram is clear or not. tomorrow i will post another pdf of the linkage with the ram fully closed (i.e. when it is at 800mm closed centres) to maybe illustrate better what i mean.

this link will in essence be trying to rotate a mass of 4tonne which will be pivotal around the main large pin.

so im trying to figure will ram force be sufficient to allow this linkage to apply a sufficient torque arm to do this.

 

RE: Linkage Torque/calculation

The key is to take the leftmost pivot where three links intersect and place it in equilibrium (i.e.the vector sum of the forces there is zero), with all 3 links having forces along their axes (the circular link is made kinematically straight as suggested).
From this you should get the torque delivered to the main shaft which is the the force in the almost vertical link x the crank arm to the main output shaft x the sine of the angle between the 2 arms. You get this by resolving the almost vertical link into 2 vectors, one along the crank link and one normal to it, the one along is the inline  crank force and the other the force which  part of the couple that makes the torque or F*sine @.   

RE: Linkage Torque/calculation

Quote:

lengthy revision of free body diagrams

Dammm! Gotta read it all again I suppose. I hope they marked them...

Regards,

Mike

RE: Linkage Torque/calculation

(OP)
Here is another PDF showing the linkage fully  closed and the linkage fully open. hopefully giving a better understanding of the dynamics involved and how it operates.

i will try what you said zekeman and see can i do this.

btw the max retraction ram force is 53KN, the mass the main pin will be rotating is 4000Kg ( my conveyor)

thanks

 

RE: Linkage Torque/calculation

(OP)
looking at this for some time now, but cant get my head around this, never was much good at the free body diagrams!

do the reaction forces at the top 2 fixed pivot points nt add a number of unknowns to our problem?

is 1 force and the link lengths etc enough to solve this?



 

RE: Linkage Torque/calculation

(OP)
I have sketched up roughly what i think the FBD for the full system looks like. i have added the reactions at the fixed points b AND c, as im sure these are important to consider. ill assume the links are of negligible weight also.


could any one guide me from here.... im not sure how to resolve this as i think i have too many unknowns.

i have added a PDF of how i think the FBD would look.

all help greatly appreciated guys.

RE: Linkage Torque/calculation

Quote:

gmcgrory, I think if you review your statics, you will find that the moment on the large pin at any position is the product of the force applied by the actuator and the perpendicular distance pin-to-actuator. Neglecting losses.

I agree with SnTMan, but I don't think you can neglect the losses. The tricky part is, of course, nailing down the geometry and I don't know if you can get away with just a static analysis.

RE: Linkage Torque/calculation

(OP)
Occupant, u think that it is just as simple as multiplying the cylinder force by the perpindicular distance from the central pivot point at different positions of the linkage?

im now in a situation where i dont know what road to go down with this tricky problem.

i hope some one can shed some light on this and help me nail it down

RE: Linkage Torque/calculation

i think you're on the right track.

let's start with your FBD.  look at the linear actuator, and the two links that attach to it.  set up a co-ordinate at end D, x-axial and y-transverse.  the two other links, AD and CD, balance this load.  as a simple example, assume AD is perpendicular to DE (the rod).  then CDcos(theta) = 53kN and AD = CDsin(theta).  in reality you'll have components from both CD and AD in both of the axes at D ... two unknowns, two equations ... GTG.

now i think 53kN is a guess (it's your maximum load, right?, but you're asking yourself waht load is it ? damnit, i'll put in the maximum and see what happens).  a good enough way to start, but at the end of the day you'll need the right answer.  the good thing is everything is linear (if the rod load 1/2s, the loads in CD and AD 1/2 as well).  the rod is reacting torque about B (yes?).  the applied load (4000kg) is through A ?, is also creating torque about B (weight?).  so you can see how to figure out the rod load to react the applied load, yes?

RE: Linkage Torque/calculation

(OP)
i basically chose this ram as i had a determined space to fit my mechanism within. this hydraulic ram with a 6oomm stroke allowed me to rotate this linkage through my required 180degrees.

the maximum retraction force this ram can provide is the 53KN i have illustrated (just a starting point i guess).

the whole linkage in essence is trying to rotate the pin B which is attached to a turntable with 4000kg on top.

i am almost understanding this now rb1957, however, when u say  "set up a co-ordinate at end D, x-axial and y-transverse", what exactly do you mean?

sorry for so many questions, but im keen to get to the bottom of this challenging problem. once i solve this, ill be alot more confident doing something similar in the future

RE: Linkage Torque/calculation

you need to determine the forces acting at D.  you know the rod force (even if it's an assumption for now and you'll calculate the correct value later), so you have 2 unknowns, the load in rods AD and CD.  you can sum forces in two directions (x- and y-) to put the point in equilibrium.  you can use any set of axes, aligning them to the rod direction makes things a little easier; using you geometry axes would work just as well  

RE: Linkage Torque/calculation

Can you put a sheet here with the linkages fully defined?

RE: Linkage Torque/calculation

show the load (ie the force due to 4000kg) on your FBD

RE: Linkage Torque/calculation

(OP)
its hard to show the load of 4000kg rb, as its basically sitting on top of the link B. the actuator is trying to rotate the link B which goes through a turntable with a mass on top of it.

my linkage is in effect being opposed by a dead load sittin on link B

i have attached a DWG file for reference guys

many thanks

RE: Linkage Torque/calculation

This may help.  Find shaft torque in terms of the link force.  Find the link force in terms of the cylinder force and link geometry.  The two links and hydraulic cylinder are all two-force links.  By the link geometry the force directions are known.  The cylinder force is known.  Solve the force triangle at the pin common to the three links.  Law of Sins relates forces in the triangle to the angles between the forces/links.  Use geometry to solve for various angles.
See my attached sketch.

Ted

RE: Linkage Torque/calculation

hi gmcgrory

Can't see attachment, however your FBD in the other one looks okay, except it looks like you have some very acute angles in that mechanism, does it actually work? its looks like the links might bind up to me.
The torque on pin marked B will be the force will be the force in the link A-D multiplied by the 132.9?? dimension.

desertfox

RE: Linkage Torque/calculation

Can't open your drawing, have only 2006.

RE: Linkage Torque/calculation

(OP)
desert fox, the straight link is actually in reality a curved link, but a few on here said i should make it a straight link for calculation purposes. so it doesnt actually bind up.

i have been trying to calculate this with all the help, but i still cannot come up with a resonable force in the link AD.

has anyone had a go at actually calculating this to see what force is present in link AD? i am coming up with a wild force of  865KN.

this cannot be correct i dont think

RE: Linkage Torque/calculation

assume for simplicity that AD is perpendicular to the actuator rod.  then the load in CD = 53kN/cos(26.73deg) = 59.34kN and the load in AD is 59.34kN*sin(26.73deg) = 26.7kN.  check 26.7^2+53^2 = 59.34^2.

RE: Linkage Torque/calculation

Hi

please reattach as pdf as well I'm happy to have a go calculating it.

desertfox

RE: Linkage Torque/calculation

full equations for ptD ...
theta = angle between CD and AD
53kn = CD*cos(26.73)+DA*cos(26.73+theta)
0 = CD*sin(26.73)+DA*sin(26.73+theta)

simplified example, theta = 90-26.73 = 63.27deg

53kN = CD*cos(26.73)+DA*cos(90)
0 = CD*sin(26.73)*DA*sin(90)

note CD is in compression, and DA is in tension

RE: Linkage Torque/calculation

As an observation, you will not get 180 degrees rotation of the rotation shaft and still have torque to drive it.  When the line of action of the driving link force passes through the shaft rotation center, input torque is zero.  It dead-centers.  Top or bottom.
Inertia may carry the drive through.  Or an over-running load may push it through.

Ted

RE: Linkage Torque/calculation

Please disregard my observation comment.  It is incorrect for the excavator linkage.  You can get 180 degree rotation.
My error.

Ted

RE: Linkage Torque/calculation

(OP)
guys, i will be back at work on monday and post a pdf as requested FAO desert fox.

all help appreciated thus far. great forum where minds can be brought together to solve engineering problems.

RE: Linkage Torque/calculation

Here is suggestion for a much simpler way to determine whether the mechanism will do what you want.  It's called the method of virtual work.  

In your drawing, attach the object and arm in place of the backhoe bucket. Then move the cylinder a SMALL amount, say 10 mm.  Find the amount the center of the weight moves lets say the weight moves 52 mm.  For that small increment, 52 X weight = 10 X cylinder force, pretty simple.  (Obviously you will need the cylinder area and available pressure to get the maximum cylinder force available.)  Then continue the increments thru the full range of desired motion.

Another method which is even simpler, although not very elegant, is to fire up the backhoe machine, clamp a piece of wood to the bucket, with a target in place of the intended 4000 Kg.  With a measuring tape, measure the vertical location of the target as a function of the cylinder displacement.  Again, use the information in increments.  

If you want to get fancy, put in two targets, one for the 4000 Kg weight, and another for the center of gravity of the arm you are going to add...then sum both the arm work and the weight work.

With either of the above methods, you need to deduct a reasonable amount for pin friction, say 10 to 20 per cent.

The above method works for backhoe mechanisms, scissor lifts, dump trucks, articulated cranes, etc.

RE: Linkage Torque/calculation

From your Pdf I've come up with AD->F=28.62kN and M=3.816N-m, CD->F=62.26kN in the shown endpoint position. Just for comparison.

RE: Linkage Torque/calculation

(OP)
bradley rather than using the linkage above to curl a bucket, i am turning it on its flat in order to rotate a mass of 4000kg using the main pin B. is there not a simple method or calculation to determine what torque is needed to rotate a certain mass about a point? is it not an inertial calculation maybe?

RE: Linkage Torque/calculation

(OP)
occupant, is M that you stated meant to be 3.816KNm? or Nm  

RE: Linkage Torque/calculation

(OP)
Nice artcicle ted...  just shows, linkages are a whole science in their own respect. smart minds initially thought of such stuff.

So now that we have more or less sorted the problem of calculating the force in the link AD, therefore multiplied by the perpindicular distnce to the pin AB gives us the max torque availale with our chosen cylinder. how can we determine if this will allow us to easily rotate our 4000kg turntable load which acts at the main pin B?

RE: Linkage Torque/calculation

Hi gmcgrory

To find out whether you can rotate the 4000kg mass, we need to first fully understand how it attached to pin B, is it sat a distance from pivot B thereby creating torque? is it in fact attached to the pin A? thats what we need to know to help you further, also we need more dimensions of the pivot centres, link lengths etc, we can't analyse the mechanism as the information is incomplete.
Also to determine the maximum torque on pivot B the mechanism needs to be stepped through a number of positions and a force analysis done in each position.
Example:- rotate the link A-B through a angle of say 2 degrees,draw all the positions of the connecting links and cylinder for this new position, do a force analysis for all the links including the force in link A-D and determine the torque on pivot B, then move the link A-B another 2 degrees and repeat the force analysis, keep doing this till you reach the final position for the mechanism, this will give you the minimum and maximum torques on pivot B throughout the mechanism movement.
I'll try and post an example later.

desertfox

RE: Linkage Torque/calculation

Quote:

occupant, is M that you stated meant to be 3.816KNm? or Nm
   
Yes, should read kN-m. By the way, the equivalent numbers for the 0.0 degree position are: A-D->F=77.85kN, M=12.12kN-m and C-D->22.29kN

RE: Linkage Torque/calculation

gmc,

Thanks for the clarification about how you intend to use the mechanism for slewing rather than lifting....

You can use the virtual work method to equate cylinder displacement times force to torque times angle.  Use angles in radians, then be sure your units are consistant for work.

By one of the methods presented in all the responces you should be able to find the available torque from the cylinder and mechanism.


Several bigger problems:

1)  How much torque is necessary to slew the conveyor?  Add up unbalanced wind load, vertical slewing axis not true vertical, friction in the pivots, and a bit to accelerate the conveyor.  If you have trouble arriving at a reasonable value, take the worst case unbalanced vertical moment and multiply by a reasonable factor, say 20 or 30 per cent.  Keep in mind that a conveyor will at one time be completely full of material, and the drive will quit, then people will shovel the product off the side to get it going again...  (from many years of experience.)

2)  Will the backhoe arm and mechanism be strong enough to carry the conveyor?  For a start, find the worst case of unbalanced vertical plane moment.  Then compare it to the section modulus of the backhoe boom times its strength.  If you dont know the materials of the boom, assume mild steel.

3)  Backhoes are designed to curl the bucket in a few seconds.  You will need to slow the slewing motion by providing a very small hydraulic flow.

Again, I am making an assumption....that the backhoe mechanism is "carrying" the conveyor.  If the conveyor is suspended by a separate king post and bearing, you can set aside the comment  number (2) about backhoe arm strength.

RE: Linkage Torque/calculation

(OP)
the turn table will in fact be attached to the pin at B through a keyed arrangement to transmit the torque. it is on this turntable that the 4000kg conveyor will be mounted.

I dont know if you have seen the DWG file i attached desertfox. i will attach a better PDF tomorrow with more measurements.

i thought it would be easier to determine what torque we would need to get the mass to start rotating........

 

RE: Linkage Torque/calculation

(OP)
looking at rb1957 equations for point D, they seem logical and correct to me. working them out for when our cylinder is at full stroke of 1400mm:

approximately, there is a force of 26KN in AD in tension and a force of 58KN in CD in compression.

as a few have said i would have to carry out this analysis for a number of positions of the linkage to determine max and min forces in AD there fore max and min torques available at the pin B.

 

RE: Linkage Torque/calculation

(OP)
when the linkage is closed and the ram is now pushing out, i calculate that AD is nw in compression at 80KN and CD is in compression at 36KN.

interested to see what others get.

RE: Linkage Torque/calculation

(OP)
CD in tension sorry, when ram is pushing out from fully closed position

RE: Linkage Torque/calculation

hi gmcgrory

I get 37.47kN and 80.91kN in the respective links where you get 36 and 80 so were close, however thats based on 53KN from the cylinder, i haven't seen yet how 53KN will be sufficient to cope with the 4000kg mass, what is the torque created on the links from the actual load your trying to move?

desertfox

RE: Linkage Torque/calculation

(OP)
"what is the torque created on the links from the actual load your trying to move?"

Desertfox, so we now have determined how to find out the forces in the links AD and cd, there fore multiplying the AD link foce by the perpindicular distance to the link B will give us the torque at various angles.

the question nw is, hw do i simply determine what is the resisting torque we must over come to allow us to rotate the turntable attached at pin B. bear in mind the 4000kg conveyor is nt acting directly through the pin, its centre of gravity will be acting at a distance from the pin. due to it acting at a distance, i belive the inertial force we need to overcome is greater.

RE: Linkage Torque/calculation

gmcgory,
Did you calculate the forces or derive them graphically?

Will the axis of pinB be vertical?
Is the rotation axis of the turntable the same as the axis of pinB?
How far is the 4000kg from the rotation axis?  Do you lift the 4000kg against gravity or move it in a plane perpendicular to gravity?
How fast do you want to accelerate?

Sketch?

 

Ted

RE: Linkage Torque/calculation

(OP)
i calculated the forces using the geometry hydtools as advised by rb1957. i made sure i fully understood the mechanics again before just simply applying the equations.

i have attached a pdf of the proposed idea. i will be accelerating at a slow rate, im not sure exactly what. it wont be 1mm/sec nor it wont be 10m/s..... just a reasonable rate at which a radial conveyor would normally be turned.

RE: Linkage Torque/calculation

In looking at your drawing, you seem to have a tensile support cable or chain attached to the boom inboard of the center of gravity of the boom.  At the end of a run (when load material is on the outboard end of the conveyor, but not on the inboard end) it seems like this arrangement could create considerable uplift on the turntable bearing, possibly requiring additional bearings.

RE: Linkage Torque/calculation

(OP)
this is really only a provisional drawing. just illustrating the idea. the chain would just take the load of the conveyor when it is in its working position rather than letting the rams hold the load.

Could you maybe explain what you mean further by it creating uplift?

Potteryshard, do you have any idea how i easily calculate what torque is required to turn this arrangement with my linkage?

RE: Linkage Torque/calculation

Hi gmcgrory

I'm having difficultly relating the orientation of your mechanism with the device your trying to move.
The best way to solve the problem is increment your load in stages ie step through the load positions and find the greatest torque or force position it exerts on your mechanism, then draw your linkages for that point and resolve the forces in them, it seems to me your doing the calculations the wrong way round.


desertfox

RE: Linkage Torque/calculation

(OP)
the main pin going thorugh the turn table, this is the pin that linkage will be operating on. i.e Pin B in the linkage we have been looking at is the the pin that the turntable and conveyor rotate around.if you can imagine the boom/conveyor  rotating +-90deg about the Z axis.

i am nw having difficulty determining what torque is needed to turn pin B. and yes, i have worked backwards really to understand how i work out what torque such a linkage was capable of with the ram i chose as a starting point.

So im at a point now where i really need to know how to determine what torque is required to turn a a load which overhangs its axis of rotation

RE: Linkage Torque/calculation

The way I understand this is you want to rotate this conveyor hanging on pin B through 180 degrees. That requires to know the inertia of the conveyor. Then you can calculate the angular acceleration anlogous to F=m*a and develop a velocity profile for the move. However, the term "velocity profile" assumes that you have some kind of servo control on your hydraulic system. I don't think a bang-bang approach would work here.

RE: Linkage Torque/calculation

gmcgrory, I'll take a shot at this.
You must first overcome friction of the pin bearing/bushing and turntable bearing.
The pinB reacts to the 4000kg, W, located distance R, 3.65m, from the pinB axis.  The pinB reaction is W*R/d, d being the assumed distance <= length of the pin or length of the pin bushing over which the pinB reactions act.  This pin reaction, fb, is the normal force used to calculate the frictional force and the rotation resistance of the pin/bushing.  The friction force is fb*mu.  Its torque of resistance is fb*mu*(bushing diameter).
If the turntable is supported by a bearing, then figure the friction force by W*mu and estimate the diameter at which this force acts to find the torque resistance, W*mu*D.
The sum of these two friction torques must be overcome by the linkage torque to get the rotation started.

Once started the friction torque may be neglected.  The weight, W, is then accelerated at some rotational acceleration, alpha.  (W/g)*alpha*R = M, the torque required to drive the swing movement.  By kinematic equations, alpha = 2*theta/t^2, where theta is the angle in radians of the swing and t is the time in seconds to move through that angle and initial velocity is 0.

Then, the torque to swing is M = (W/g)*R*alpha = (W/g)*R*(2*theta/t^2)

Certainly open to critique.

Ted

RE: Linkage Torque/calculation

Hi gmcgrory

Okay I think I get the picture, now I agree with others about friction and inertia etc, however if your moving this thing really slow you can usually ignore the dynamic aspects,then your static analysis might just be enough of course you need to build in margin on your cylinder to make sure it does move.
So maximum torque on pin B would be 4000kg * max lever arm length, then draw your load in various positions throughout the mechanism movement and again draw link positions and calculate forces.

desertfox

RE: Linkage Torque/calculation

as a sidenote...
this very mechanism was used in some early washing machines

The main shaft being the agitator in the tub (can't say for sure if there was more than 180deg rotation in the one I tore apart as a kid)

The long link (shown as the actuator in your example) was an arm attached to a crank driven by a reduction gear from the elect motor.

All of the mecahnism was inside a cast alum gearbox.
It was intriguing to see how it worked.

RE: Linkage Torque/calculation

(OP)
so say the max lever arm is as i said it was 3.65m and the mass is 4000kg, then the torque on pin is 4000 x 9.81 x 3.65 Nm.

 is this the max torque i need to provide using the cylinder to rotate the conveyor? i should add 20-30% for friction plus margin for error i guess?

 

RE: Linkage Torque/calculation

Yes, that sort of approach I would agree with. And yes, your lever arm would be at maximum torque at 3.65m.
desertfox

RE: Linkage Torque/calculation

No.  4000*9.81*3.65 would be the moment applied to tilting/prying/bending the pinB in its bushing longitudinally to the pinB axis, not twisting the pinB around its axis.  The pinB and its bushing must resist being bent by this moment.  This is not the moment against which you must apply torque through the pinB to move the conveyor.
As I understand your conveyor arrangement, it swings in a horizontal plane, not a vertical plane.  The pinB axis is vertical, not horizontal.  Or is my understanding of your conveyor arrangement incorrect?

Ted

RE: Linkage Torque/calculation

(OP)
hydtools, ur understanding is correct, the conveyor does indeed swing in the horizontal plane. what do you suggest knowing this?

RE: Linkage Torque/calculation

I thought it was in the vertical plane, so hydtools is correct the moment you need to overcome is that due to fricton from the 3.65m * 4000*9.81 and because the pin axis is at 90 degrees to the movement your pin will have to resist bending and shear.

desertfox

RE: Linkage Torque/calculation

A problem well stated is a problem half solved:)

RE: Linkage Torque/calculation

gmcgrory,
Estimates and guesses:
PinB diameter = 100mm
PinB bush length = 400mm
PinB reaction forces separated by 340mm
friction coefficient = .16
PinB friction torque = 4000*9.81*3.65*(1000/340)*.16*(100/1000) = 6740Nm
Turntable bearing is ring shaped:
150mm id, 300mm od, average radius 112.5mm
friction coefficient .16
Turntable friction torque = 4000*9.81*.16*(112.5/1000) = 706Nm
Total static friction torque = 6740 + 706 = 7446Nm

Inertia torque:
guess 45deg rotation in 3 secs.; theta = .785rad
4000*3.65*2*.785/9 = 2547Nm

For order of magnitude estimate.

Ted

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