Abdominal Pressure


I am working with the spine of the FreePostureMove model. I eliminated arms and legs. I am trying to understand the abdominal pressure. I believe the pressure is applied by the Artificial Support Muscle, is this correct?
When I run the model with no external loads but gravity and no movement I get Fm=48.1N at Disc 3 and Fm=45.7N at Disc 4. Was this intended? Are there any refs supporting abdominal pressure in neural standing?
Also, if I move the COR of the joint,(by changing the srel values at L5Seg L4L5JointNode and L4Seg L4L5JointNode) I notice that these values change. How can I move the COR of the joint without causing change in these values?
Finally I tried turning off the abdominal support, using the BuckleNoAbdominalPressure.any file, but it does not run because it is missing Cyl.any file. Can you guys post that file?

Thanks in advance for your help,

Lissette Ruberte

Hi Lissette

The abdominal pressure model is rather complicated and difficult to explain in detail.

The best description you will find at this link,


please be very patient, the file is rather large, please look at slide 48 and forward.

Basically the abdominal pressure model works like a “mechanical” ballon, the transversus muscles are squeezing it and it can produce an extension moment on the spine. The abdominal pressure muscle will only be active if it of benefit for the other muscles and it is recruited like any other muscle.

The forces you are reporting are some internal quantities that is usually not of interest, basically these values shows how much each of the transversus muscles press against the volume.

I have tried to replicate the numbers you provide but i have not been able to do this. I took the FreePostureMove model from the new AMMRV1 repository excluded the legs and arms, removed gravity, and added muscles in the spine. When running this model i have zero activity and and zero force in the support muscles you list. Could it possibly be that you have been running the model with some motion? this will produce some forces due to centrifugal forces…

Concerning the missing cyl.any file i am afraid that you have tried to run the root file BuckleNoAbdominalPressure.root.any file. This file is a leftover in the repository and should have been removed from the directory earlier. The file relates to very old version of the abdominal model, sorry for this.

It is not so easy to remove the abdominal model because many of the muscles are connected to the buckle and if the abdominal muscle is removed it may lack support and make the solution fail. In the bottom of the Trunk/Buckle.any file you will find the Abdominal muscle and if the strength of this is set to zero it should remove the effect of the abdominal pressure, but the downsize is that it may introduce instability into the solution.

Concerning moving the joints: When you move the sRel values of the joints it will change the kinematics and kinetics of the model so any change in the kinematics will have an influence on the calculated forces, there is no way that this can be avoided.

Hope these answers helps you move on, otherwise please write again.

Best regards

HI Soren

Thanks for your response. I realized that by setting the abdominal muscle model force to zero, the solution is indeterminate. However I can find a solution if I eliminate the transversus muscle from the model. This in effect eliminates the abdominal pressure.
I decided to use the model with abdominal pressure but I am trying to measure its effect of the lumbar joints reactions. I used AnyForce MomentMeasure2. I first solved for all muscle forces about L5S1 joint without transversus muscle and found that the measured reactions were equal to my constraint reactions. However, when I add the transversus muscle I do not know how to add the abdominal pressure effect.

I added the following lines to the the measure of all muscle forces done previously without the transversus

AnyForceBase &T1=…TransversusL1;
AnyForceBase &T2=… TransversusL2;
AnyForceBase &T3=… TransversusL3;
AnyForceBase &T4= …TransversusL4;
AnyForceBase &T5=… TransversusL5;

AnyForceBase &AP=…AbdominalPressureMuscle;
AnySeg &Buckle = …Segments.BuckleSeg;
AnySeg &disc1 =…Disc1.Disc.DiscSeg;
AnySeg &disc2 =…Disc2.Disc.DiscSeg;
AnySeg &disc3 =…Disc3.Disc.DiscSeg;
AnySeg &disc4 =…Disc4.Disc.DiscSeg;
AnySeg &disc5 =…Disc5.Disc.DiscSeg;
AnyKinLinearComb &APVolumne =…Buckle.AbdominalVolume;

Include Gravity =On;
Include InertiaForce=On;

But the measured reaction does not match the constraint reactions of the joints. I have done a transformation so that all measurements are in the global coordinate system. Can the AnyForceMomentMeasure2 be used to measure the effect of abdominal pressure? If so, what am I missing to include


Hi Soren

I was using AnyForceMomentMeasure2 to measure the effect of the transversusL1 on the lumbar joint. I noticed that if I included the L1 Lumbar segment alone,the total reaction force given by AnyForceMomentMeasure2 is half the value of the muscle or Fm value. If I included the AnySeg &disc1…buckle.disc1.discseg and AnySeg &L1=…SegmentsLumbar.L1Seg, the value was double the Fm value. But if I included the just the buckle disc segment that value was slightly greater than the Fm value of the muscle.

I know that the transversus wraps around the entire body. To measure its effect what segments should one define in the AnyMomentMeasure2


Hi Lisette

It is very complicated to measure the effect of the abdominal pressure in this way with the MomentMeasure2, this comes from the complicated mechanism of the abdominal model, it requires the full overview of it and being able to select the right items to go into the measure. The ForceMomentMeasure2 basically does a free body diagram, so it is important to know what to include in the measure. So it should be possible, but it is by no means a simple task.

I think the best way to measure its effect would be to excluded the abdominal pressure and compare the two solutions. Right now there is no standard way of excluding it but the method you describe sounds ok. Another option would be to lower the abdominal muscle strength significantly, but this may have the result that this muscle would determine the muscle activity envelope.

In the wiki pages there is a small example of the use of AnyForceMomentMeasure2 maybe it can be helpful


Best regards

HI Soren

I have two questions,

  1. The transversus muscle starts and ends at the same point in a circular pattern. It terms of the effect of this muscle on the lumbar joint load, doesn’t it cancel itself out? Besides, creating intraabdominal pressure, does the transversus contribute to balance the external load?

  2. What type of reactions does the intraabdominal pressure create about the joint loads? Shear, Tension and extension moment?


Hi Lisette

You are absolutely right about the tranversus muscles their effect are only on the abdominal pressure, they can help create pressure, nothing else.

The intra abominal pressure and create an extension moment , but it also has forces acting horizontal on each of the vertebraes, these can create shear forces.

Best regards

Hi Soren,
Your explanation about abdominal pressure is very helpful. But I have a problem.
I measured all the force in L1 using AnyMechOutputFileForceExport .
But I find two force as follow. They act on same node and have the same value but opposite direction. In terms of L1, they will cancel each other. As you said, the abdominal pressure should create shear force. So I dont understand the meaning of the .Disc1.Jnt.Constraints.Reaction. The version of my Anybody is AMMR 1.1.

Pos= -0.02; 1.20; 0.00;
F[0]= 11.89; 0.60; -0.00;
M[0]= -0.00; -0.00; -0.00;

Pos= -0.02; 1.20; 0.00;
F[0]= -11.89; -0.41; 0.00;
M[0]= -0.00; 0.00; 0.00;
F[1]= 0.01; -0.19; -0.00;
M[1]= -0.00; -0.00; -0.00;
F[2]= -0.00; -0.00; -0.00;
M[2]= -0.00; 0.00; 0.00;

Best regards

This force is a cumulative reaction force acting on the L1 segment. As you see it does not completely cancel the transversus muscle force. If you try to export the force information on other segments, say L3, you will notice the similar behaviour, except that the other components will have larger magnitudes due to the incline of the vertebral bodies.

Kind regards,


Hi Pavel,

I encounter the same problem. And I export all forces acting on L3 for 40 degree flexion. However, when you compare these two forces, you will find that the total magnitude is totally the same(0.6=0.41+0.19). There is no shear force acting on this joint. Can you confirm this? These two forces cancel out each other in Disc plane. Does this mean there is no shear force produced by abdominal pressure? Or I did something wrong?