Northtowne Chiropractic
Dr. Larry Van Such, DC, BE
Dr. Rick Horsfield, DC
Jamie Ike, LMT

2240 Morse Road
Columbus, Ohio 43229

Office Hours
Mon & Wed   9am - 12pm
  3pm - 5pm
Tue & Thur   12pm - 5pm
Fri   9am - 3pm
Sat   By Appt Only

(614) 428-9310

Dr. Larry Van Such

Lifting Heavy Object; Bending and Twisting with 80 lbs weight.

This is an actual report that was submitted to the Ohio BWC Industrial Commision on behalf of my patient.  The employer's attorney argued that it was impossible to cause a disc injury with this mechanism. The only problem is that he didn't have any professional background to justify this statement nor could he find any other professionals to support this claim other than hand picked Independent Medical Doctors hired by the employer who could only offer their opinions based on "feelings" and not actual engineering mechanics.  Had I not written this report, he would have gotten away with it and the allowance would have been denied.  This claim is now allowed for a central disc protrusion at L5-S1 and the patient is now receiving proper care.  I also show up to hearings to ensure this report is heard and typically do the same for all cases that I believe have merit.

I would be happy to do the same for you if I believe your injury caused more than just a contusion or sprain.  Call me to discuss your situation.

Dr. Larry Van Such, DC, BE, EE
2240 Morse Road
Columbus, Ohio  43229

Phone: (614) 428-9310  Fax:  (614) 428-9407

August 24th, 2010
RE:  Michael
Claim No.:  10-xxxxxx


                Lifting Heavy Object


                Central/Left Paracentral Disc Protrusion at L5-S1

To whom it may concern:

Michael xxxxx injured his lower back on 0X/XX/2010.  The mechanism of injury as presented by the Michael for this injury is as follows:  Michael was lifting pans as they are called of steel ceiling tiles.  Each weighed between 80-120 lbs. depending upon the grade of steel that was being used.  Michael had to grab a box of these tiles by wrapping his arms around the box.  He held it close to his body and as he was twisting and lowering it to the ground to place on a skid, it started to slip from his grip.  He held on to it as it carried him to the floor.  This is when Michael felt severe pain in his lower and middle back.

During this lifting, twisting and lowering of the weight to the ground, Michael was not able to initially bend his knees and lower the weight properly because the table in which the weight was on was in his way.  He eventually ended up in a position where is upper body was bent over at the waist at an approximate angle of  60º when the injury happened.

In order to calculate the amount of force this mechanism could deliver to his spine from a minimum of 80 lbs.  originating near his chest, we need to measure how far the center of gravity this object was from the injured L5/S1 level.  From the description Michael gave, this estimated distance was 12 inches.  We also need to determine the angle this force would be had it been perpendicular to his spine and not perpendicular to the floor.  This angle is 30º and it is needed to calculate the moment of force, and ultimately the total force as described on the next page.

From the above description, we can construct a free body diagram of this mechanism.

Free body diagram:

Total F1 Force acting upon L5/S1:

First we need to calculate the moment of force.  The moment of force supplied to the L5/S1 level while attempting to lift an 80 lb. object is calculated as follows:

We start by converting the 80 lb. force at 30º to the spine to one that is positioned 90º to the spine.  Therefore:

                F1 @ 90º = 80 lb. x’s cosine (30º) = 69.30 lb.

The magnitude of this moment of force is equal to the perpendicular force times the distance d.  The distance “d” is referred to as the moment arm or perpendicular distance from the point of interest to the line of action of the force.  Therefore:

                MF1 = F1 @ 90º  x’s  Distance = 69.30 lb. x’s 1 ft = 69.30 lb. ft.

                Converting 69.30 lb. ft. to lb. inches:

                69.30 ft x’s (12 inches / 1 ft.) = 831.38 lb. inches.

                MF1 @ L5/S1  = 831.38 lb. inches.

Now that we know the moment of force F1, we can determine the specific force at the L5/S1 level.  The estimated and average diameter of a lumbar disc is 2 inches.  Therefore, the approximated force generated on L5/S1 disc is:

                F1@ L5/S1    = (831.38 lb. inches) / (2 inches) = 415.69 lbs.

This 415.69 lb. force is the conservative estimate that would have been applied to Michael’s lower back during the injury since we used the lower range of weight instead of the higher (120 lbs).  However, this does not take into account the normal stress and strain that the spine undergoes when there is no additional load applied.

Total F2 Force acting upon L5/S1:

Under normal situations with no additional load, the lumbar spine undergoes strain due to the inherent body weight of the individual.  Therefore, we first estimate 1/2 of the patient’s body weight, which is the weight from the waist of the patient up to their head.  Then, we estimate the center of gravity would be near the sternum which is located approximately at a perpendicular distance of 12 inches from the L5/S1 level:

Michael’s total weight is approximately 228 lbs.; half of this is 114 lbs., which is a reasonable estimate of weight from his waist up to his head.  114 lbs. of weight, with a center of gravity near the sternum will generate a moment of force at the L5/S1 disc level as follows:

Again, we start by converting the 114 lb. force at 60º to the spine to one that is positioned 90º to the spine. 

                F2 @ 90º = 114 lb. x’s cosine (90  -  60º) = 98.75 lb.  (note: 30º is the angle used to bring it to perpendicular)

Also, the magnitude of this moment of force is equal to the perpendicular force times the distance d.  The distance “d”, as stated above, is referred to as the moment arm or perpendicular distance from the point of interest to the line of action of the force.  In this case, d = 12 inches or 1 ft.  Therefore:

                MF2 = F2 @ 90º  x’s  Distance = 98.75 lb. x’s 1 ft. = 98.75 lb. ft.

                Converting  lb. ft. to lb. inches:

                98.75 lb. ft x’s (12 inches / 1 ft.) = 1185.03 lb inches.

                MF2 @ L5/S1  = 1185.03 lb inches.

Now that we know the moment of force F2, we can determine the specific force at the L5/S1 level.  The estimated and average diameter of a lumbar disc is 2 inches.  Therefore, the approximated force generated on L5/S1 disc is:

                F2@ L5/S1    = (1185.03 lb. inches) / (2 inches) = 592.52 lbs.

Calculating Total Force at the L5/S1 level from lifting an 80 lb. weight:

                FTotal @ L5/S1 = F1 + F2

                FTotal @ L5/S1 = 415.69 lbs. +  592.52 lb. = 1,008.21 lbs or 0.50 Ton.

This is the maximum total force that could have been applied at the L5/S1 level in Michael xxxxx’s spine.  This number is consistent with previous studies that indicate that forces in the order of ten times the lifted load can be easily generated in the lumbar spine.  (80 x 10 = 800 lbs.) To gain a realistic appreciation of whether or not this amount of force can be reasonably generated, consider the amount of torque one is able to administer when manually removing lug nuts off a tire.  Torque can be generated up to 1200 lbs. with the use of a tire iron simply by using the arms.  In this case, the tire iron serves as the moment arm similar to the spine.

Protective mechanisms: 

If any of the body’s protective mechanisms are inhibited, then any amount of force, even as low as 5 lbs, acting at a specific level in the spine can cause significant and lasting damage.  Protective mechanisms would include the body’s supporting musculature guarding the spine.  If muscles give out due to the load, they will tear and subject the spine to serious injury.  Ligaments then bear the majority of force, become exposed and tear further subjecting the spine to injury.  If that occurs, then the disc will be exposed. 

Since it has already been established that Michael was diagnosed with a lumbosacral sprain, it is reasonable to assume that his disc was exposed and subjected to this force.

Intervertebral joint motion:  The intervertebral joint (one disc and two facet joints) is a type of universal joint with six degrees of freedom;  three of translation (sideways, backwards-and-forwards, and up-and-down) and three of rotation (side-to-side), bending forwards and backwards, and longitudinal rotation).  Typically, however, the joint does not perform simple movements in isolation, but always couples one translational with one rotational movement, such as forward bending and left rotation.  If these coupled movements occur out of the normal range of motion for which a patient is unable to remain in control of their movements, muscles weaken and expose the patient to serious injury.


Michael’s mechanism of injury would easily have subjected him to coupling motions in his lumbar spine beyond his control resulting in injury to his spine.

Therefore, it is my professional opinion, and within a reasonable degree of chiropractic and engineering mechanics certainty, that Michael xxxxxx’s mechanism of injury was significant enough to cause a central/Left Paracentral Disc Protrusion at L5-S1 at the time of the injury.  This additional finding is consistent with all of his clinical findings and should be added to his claim.

Respectfully submitted,



Dr. Larry Van Such, DC, BE, EE
Doctor of Chiropractic, 1994-Present
Bachelor of Engineering, Youngstown State University, 1986
Major: Electrical Engineering
Senior Controls Engineer:  United Technologies Corporation
   Pratt & Whitney Aircraft: 1986-1991

Text:  Engineering Mechanics-Statics
                Hibbeler, 3rd Edition

Dept. of Civil/Environmental & Chemical Engineering
Youngstown State University, Youngstown, Ohio.

Journal of the Royal Society of Medicine
Volume 79, February 1986
Pages 100-104.