`______________________
`
`BEFORE THE PATENT TRIAL AND APPEAL BOARD
`______________________
`
`MEDTRONIC, INC., MEDTRONIC VASCULAR, INC.,
`and MEDTRONIC COREVALVE, LLC
`Petitioner
`
`v.
`
`TROY R. NORRED, M.D.
`Patent Owner
`______________________
`
`Case IPR2014-00110
`Patent 6,482,228
`______________________
`
`DECLARATION OF CARL T. RUTLEDGE, Ph.D.
`
`NORRED EXHIBIT 2097 - Page 1
`Medtronic, Inc., Medtronic Vascular, Inc.,
`& Medtronic Corevalve, LLC
`v. Troy R. Norred, M.D.
`Case IPR2014-00110
`
`
`
`I, Carl T. Rutledge, declare as follows:
`
`1.
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`I am a citizen of the United States and a resident of Ada,
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`Oklahoma. My post office address is 2009 East 11th Street, Ada, Oklahoma 74820-
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`7006.
`
`2.
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`In 1966, I received a B.S. with honors in Physics and Mathematics
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`form the University of Arkansas (Fayetteville). In 1969, I received a M.S. in Physics
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`from the University of Arkansas (Fayetteville). In 1971, I received a Ph.D. in Physics
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`from the University of Arkansas (Fayetteville).
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`3.
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`I am currently employed as a Linscheid Distinguished Professor
`
`and Chairman in the Physics Department at East Central University in Ada,
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`Oklahoma.
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`4.
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`From 1981 until the present, I have been employed full-time as a
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`Professor in the Physics Department at East Central University, where I have also
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`taught Astronomy, Computer Science and Mathematics. From 1970 to 1981, I was
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`employed as a Professor of Physics at Southern Arkansas University (Magnolia).
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`5.
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`6.
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`A copy of my current Curriculum Vitae is filed as Exhibit 2092.
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`I have been asked to review the work of Dr. Stephen Lombardo
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`and assist in explaining the equations used in his mathematical model. I understand
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`the purpose of the model was to confirm that a prosthetic aortic valve could be held
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`in place by a stent that was within the size range tolerated by the adult human
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`NORRED EXHIBIT 2097 - Page 2
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`
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`anatomy without rupturing the aorta using mainly the frictional forces between the
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`stent and the aorta
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`7.
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`In order to perform my work, I was provided with an electronic
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`copy of Dr. Lombardo’s model, Exhibit 2084, as well as a copy of his handwritten
`
`notes, Exhibit 2019.
`
`8.
`
`Referring to Exhibit 2109, pressure is given in mm Hg, or
`
`millimeters of mercury. Standard air pressure at sea level will push a mercury column
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`up a tube evacuated at the top to a height of 760 mm = 76 cm. (Standard mercury
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`barometers operate on this principle.) 760 mm is called one atmosphere and
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`corresponds in US units to 14.7 pounds/square inch, or 14.7 lb/in2. This means 1
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`lb/in2 = 760 mm Hg/14.7 = 51.7 mm Hg.
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`9.
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`Maximum pressure to be exerted on the aorta is 300 mm Hg, or
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`30 cm Hg, as 10 mm = 1 cm, which is the most pressure a normal heart can supply.
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`10.
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`An aortic rupture (bursting) pressure of 1200 mm or 120 cm Hg
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`was used, which greatly exceeds the rupture data provided by Dr. Norred.
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`11.
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`The coefficient of friction (cof) is found by dividing the tangential
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`sliding force Ft parallel to the surface by the normal force Fn pushing down on the
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`surface:
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`cof = Ft/Fn
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`NORRED EXHIBIT 2097 - Page 3
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`12.
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`The larger the coefficient of friction, the harder it is to slide one
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`object over another. For example, if it takes 10 lb (Ft) of horizontal force to slide a 10
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`lb (Fn) box across the floor, the coefficient of friction is 10/10 = 1 and the box is
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`hard to slide. If it takes only 1 lb of force to slide the box, the coefficient of friction is
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`only 1/10 or 0.1 and the box is 10 times easier to slide.
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`10 lb
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`10 lb
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`1 lb
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`10 lb
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`cof = 10 lb/10 lb = 1
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`cof = 1 lb/10 lb = 0.1
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`13.
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`The minimum coefficient of friction between a stent and an artery
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`wall was taken to be 0.2. Actually, since a stent will depress slightly into an artery wall
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`due to its mesh shape, the coefficient of friction probably would be significantly
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`larger, making the stent more securely locked in place. This is similar to the tread on a
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`tire giving more traction in snow than a bald tire.
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`14.
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`Aorta diameter was taken as 3.0 cm (a little over one inch, as 2.54
`
`cm = 1 inch.).
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`15. Next the amount of force needed to hold the valve in place in the
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`heart was calculated. The area of a circular valve of radius R is A = π R2, , where π =
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`3.14.
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`-4-
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`NORRED EXHIBIT 2097 - Page 4
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`Radius = R
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`The force F on the valve is pressure P times area A, or F = PA. From here on we
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`will use cm instead of mm.
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`16.
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`The force = 30 cm Hg x 3.14 x (1.5 cm)2= about 200 cm Hg cm2.
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`This is about 6 pounds in US units.
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`17.
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`The unit cm Hg cm2 is a unit of force not commonly encountered
`
`but useful in these calculations. For perspective, 1 pound = 33 cm Hg cm2.
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`18.
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`If the force of the stent on the walls is 200 cm Hg cm2, then the
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`area A of the stent must be area = force/pressure, or A = F/P = 200 cm Hg cm2/30
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`cm Hg = about 7 cm2.
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`19.
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`If we assume as a stent model sections of wire 0.1 cm width, 1.0
`
`cm length and an angle of 60° (to make equilateral triangles as shown below) and let
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`there be 20 sections of wire in each,
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`/\/\/\/\/\/\/\/\/\/\
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`then the area of each stent would be 20 pieces x 1 cm x .1 cm = 2 cm2 each, so four
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`stents such as this would be needed. These could be combined into a single stent.
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`NORRED EXHIBIT 2097 - Page 5
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`20. Next the size of a stent needed to exert a pressure of 300 mm Hg
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`on the artery walls while providing this frictional force was calculated.
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`21.
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`Since cof = Ft/ Fn, multiplying by Fn and dividing by cof yields,
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`according to algebra, Fn = Ft /cof or Fn = 200 cm Hg cm2 /0.2 = 1000 cm Hg cm2.
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`This is the force that must be exerted perpendicular to the wall of the aorta by the
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`stent to be certain the stent will not slip when 30 cm Hg of pressure is applied to the
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`attached valve, as illustrated in the following diagram:
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`22.
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`Since Fn = PA, if we set P = 140 cm Hg and use Fn = 1000 cm
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`Hg cm2, then by algebra, dividing both sides by P gives:
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`A = Fn/P = 1000 cm Hg cm2/ 140 cm Hg = about 7 cm2 .
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`23. Now 140 cm Hg is approximately the 120 cm Hg rupture stress of
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`the aorta, so the aorta would be just beyond the point of rupture under those
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`conditions. By adjusting the numbers it is easy to see how rupture can be avoided.
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`Since F = PA, multiplying A by a number and dividing P by the same number will
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`-6-
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`NORRED EXHIBIT 2097 - Page 6
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`give the same result for F. If we let the number be 4, then mathematically this is F =
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`PA = (P/4)(4A), since 4 times ¼ = 1. For example, if the area were increased 4
`
`times, the pressure exerted by the stent would be reduced 4 times. Thus the area
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`would be 4 x 7 = 28 cm2 and the pressure only 140/4 = 35 cm Hg, about the same as
`
`the 30 cm Hg maximum the aorta normally encounters and much less than the 120
`
`cm Hg rupture stress. This means that stents are a practical way to anchor the valve.
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`The pressure may be adjusted as required by increasing or decreasing the area of the
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`stent.
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`24.
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`To see how long the stent would be, consider the stent as a solid
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`cylinder, neglecting the small cuts in it because the aorta wall transmits the force
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`across them. We can find the length of a 3 cm diameter stent with an area of 28 cm2
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`using the formula for the surface area of the side of a cylinder:
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`Area = π x diameter x length, or A = πDL.
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`A = π D L
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`Length L
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`Diameter D
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`25. Dividing both sides by πD gives L = A/πD, so to get the length
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`we divide the area by the product of π and D. D is the aorta diameter of 3 cm. Then
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`L = 28 cm2/(π x 3 cm) = 2.97 cm or about 3 cm, just over an inch and a reasonable
`
`length for a stent.
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`-7-
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`NORRED EXHIBIT 2097 - Page 7
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`26.
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`An advantage of using stents as opposed to anchoring the valve at
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`a single point is that the stent spreads the total force used to hold the valve in place
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`over a large area, reducing the force on the aorta and eliminating the danger of aortic
`
`rupture. Also, if part of the aorta where the stent is located is larger than other parts,
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`less force will be exerted on that part, since the stent “spring” will be more extended,
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`again reducing the danger of rupture, as larger vessels rupture more easily than small
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`ones because there is more total force on them. (F = PA and larger vessels have
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`larger A. If the pressure remains the same, there is thus more force on a larger
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`vessel.)
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`I hereby declare under the penalty of perjury under the laws of the United States of
`
`America that the foregoing is true and accurate to the best of my knowledge and
`
`understanding.
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`-8-
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`NORRED EXHIBIT 2097 - Page 8
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