UNITED STATES PATENT AND TRADEMARK OFFICE
`______________________
`
`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 IPR2014-00395
`Patent 6,482,228
`______________________
`
`DECLARATION OF CARL T. RUTLEDGE, Ph.D.
`
`NORRED EXHIBIT 2297 - Page 1
`Medtronic, Inc., Medtronic Vascular,
`Inc., & Medtronic Corevalve, LLC
`v. Troy R. Norred, M.D.
`Case No. IPR2014-00395
`
`
`______________________
`
`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 IPR2014-00395
`Patent 6,482,228
`______________________
`
`DECLARATION OF CARL T. RUTLEDGE, Ph.D.
`
`NORRED EXHIBIT 2297 - Page 1
`Medtronic, Inc., Medtronic Vascular,
`Inc., & Medtronic Corevalve, LLC
`v. Troy R. Norred, M.D.
`Case No. IPR2014-00395
`
`
`
`I, Carl T. Rutledge, declare as follows:
`
`1.
`
`I am a citizen of the United States and a resident of Ada,
`
`Oklahoma. My post office address is 2009 East 11th Street, Ada, Oklahoma 74820-
`
`7006.
`
`2.
`
`In 1966, I received a B.S. with honors in Physics and Mathematics
`
`form the University of Arkansas (Fayetteville). In 1969, I received a M.S. in Physics
`
`from the University of Arkansas (Fayetteville). In 1971, I received a Ph.D. in Physics
`
`from the University of Arkansas (Fayetteville).
`
`3.
`
`I am currently employed as a Linscheid Distinguished Professor
`
`and Chairman in the Physics Department at East Central University in Ada,
`
`Oklahoma.
`
`4.
`
`From 1981 until the present, I have been employed full-time as a
`
`Professor in the Physics Department at East Central University, where I have also
`
`taught Astronomy, Computer Science and Mathematics. From 1970 to 1981, I was
`
`employed as a Professor of Physics at Southern Arkansas University (Magnolia).
`
`5.
`
`A copy of my current Curriculum Vitae is attached as Exhibit
`
`2298.
`
`6.
`
`I have been asked to review the work of Dr. Stephen Lombardo
`
`and assist in explaining the equations used in his mathematical model. I understand
`
`the purpose of the model was to confirm that a prosthetic aortic valve could be held
`
`NORRED EXHIBIT 2297 - Page 2
`
`
`
`in place by a stent that was within the size range tolerated by the adult human
`
`anatomy without rupturing the aorta using mainly the frictional forces between the
`
`stent and the aorta
`
`7.
`
`In order to perform my work, I was provided with an electronic
`
`copy of Dr. Lombardo’s model, Exhibit 2286, as well as a copy of his handwritten
`
`notes, Exhibit 2219.
`
`8.
`
`Referring to Exhibit 2219, pressure is given in mm Hg, or
`
`millimeters of mercury. Standard air pressure at sea level will push a mercury column
`
`up a tube evacuated at the top to a height of 760 mm = 76 cm. (Standard mercury
`
`barometers operate on this principle.) 760 mm is called one atmosphere and
`
`corresponds in US units to 14.7 pounds/square inch, or 14.7 lb/in2. This means 1
`
`lb/in2 = 760 mm Hg/14.7 = 51.7 mm Hg.
`
`9.
`
`Maximum pressure to be exerted on the aorta is 300 mm Hg, or
`
`30 cm Hg, as 10 mm = 1 cm, which is the most pressure a normal heart can supply.
`
`10.
`
`An aortic rupture (bursting) pressure of 1200 mm or 120 cm Hg
`
`was used, which greatly exceeds the rupture data provided by Dr. Norred.
`
`11.
`
`The coefficient of friction (cof) is found by dividing the tangential
`
`sliding force Ft parallel to the surface by the normal force Fn pushing down on the
`
`surface:
`
`cof = Ft/Fn
`
`-3-
`
`NORRED EXHIBIT 2297 - Page 3
`
`
`
`12.
`
`The larger the coefficient of friction, the harder it is to slide one
`
`object over another. For example, if it takes 10 lb (Ft) of horizontal force to slide a 10
`
`lb (Fn) box across the floor, the coefficient of friction is 10/10 = 1 and the box is
`
`hard to slide. If it takes only 1 lb of force to slide the box, the coefficient of friction is
`
`only 1/10 or 0.1 and the box is 10 times easier to slide.
`
`10 lb
`
`10 lb
`
`1 lb
`
`10 lb
`
`cof = 10 lb/10 lb = 1
`
`cof = 1 lb/10 lb = 0.1
`
`13.
`
`The minimum coefficient of friction between a stent and an artery
`
`wall was taken to be 0.2. Actually, since a stent will depress slightly into an artery wall
`
`due to its mesh shape, the coefficient of friction probably would be significantly
`
`larger, making the stent more securely locked in place. This is similar to the tread on a
`
`tire giving more traction in snow than a bald tire.
`
`14.
`
`Aorta diameter was taken as 3.0 cm (a little over one inch, as 2.54
`
`cm = 1 inch.).
`
`15. Next the amount of force needed to hold the valve in place in the
`
`heart was calculated. The area of a circular valve of radius R is A = π R2, , where π =
`
`3.14.
`
`-4-
`
`NORRED EXHIBIT 2297 - Page 4
`
`
`
`Radius = R
`
`The force F on the valve is pressure P times area A, or F = PA. From here on we
`
`will use cm instead of mm.
`
`16.
`
`The force = 30 cm Hg x 3.14 x (1.5 cm)2= about 200 cm Hg cm2.
`
`This is about 6 pounds in US units.
`
`17.
`
`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.
`
`18.
`
`If the force of the stent on the walls is 200 cm Hg cm2, then the
`
`area A of the stent must be area = force/pressure, or A = F/P = 200 cm Hg cm2/30
`
`cm Hg = about 7 cm2.
`
`19.
`
`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
`
`there be 20 sections of wire in each,
`
`/\/\/\/\/\/\/\/\/\/\
`
`then the area of each stent would be 20 pieces x 1 cm x .1 cm = 2 cm2 each, so four
`
`stents such as this would be needed. These could be combined into a single stent.
`
`-5-
`
`NORRED EXHIBIT 2297 - Page 5
`
`
`
`20. Next the size of a stent needed to exert a pressure of 300 mm Hg
`
`on the artery walls while providing this frictional force was calculated.
`
`21.
`
`Since cof = Ft/ Fn, multiplying by Fn and dividing by cof yields,
`
`according to algebra, Fn = Ft /cof or Fn = 200 cm Hg cm2 /0.2 = 1000 cm Hg cm2.
`
`This is the force that must be exerted perpendicular to the wall of the aorta by the
`
`stent to be certain the stent will not slip when 30 cm Hg of pressure is applied to the
`
`attached valve, as illustrated in the following diagram:
`
`22.
`
`Since Fn = PA, if we set P = 140 cm Hg and use Fn = 1000 cm
`
`Hg cm2, then by algebra, dividing both sides by P gives:
`
`A = Fn/P = 1000 cm Hg cm2/ 140 cm Hg = about 7 cm2 .
`
`23. Now 140 cm Hg is approximately the 120 cm Hg rupture stress of
`
`the aorta, so the aorta would be just beyond the point of rupture under those
`
`conditions. By adjusting the numbers it is easy to see how rupture can be avoided.
`
`Since F = PA, multiplying A by a number and dividing P by the same number will
`
`-6-
`
`NORRED EXHIBIT 2297 - Page 6
`
`
`
`give the same result for F. If we let the number be 4, then mathematically this is F =
`
`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
`
`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.
`
`The pressure may be adjusted as required by increasing or decreasing the area of the
`
`stent.
`
`24.
`
`To see how long the stent would be, consider the stent as a solid
`
`cylinder, neglecting the small cuts in it because the aorta wall transmits the force
`
`across them. We can find the length of a 3 cm diameter stent with an area of 28 cm2
`
`using the formula for the surface area of the side of a cylinder:
`
`Area = π x diameter x length, or A = πDL.
`
`A = π D L
`
`Length L
`
`Diameter D
`
`25. Dividing both sides by πD gives L = A/πD, so to get the length
`
`we divide the area by the product of π and D. D is the aorta diameter of 3 cm. Then
`
`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.
`
`-7-
`
`NORRED EXHIBIT 2297 - Page 7
`
`
`
`26.
`
`An advantage of using stents as opposed to anchoring the valve at
`
`a single point is that the stent spreads the total force used to hold the valve in place
`
`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,
`
`less force will be exerted on that part, since the stent “spring” will be more extended,
`
`again reducing the danger of rupture, as larger vessels rupture more easily than small
`
`ones because there is more total force on them. (F = PA and larger vessels have
`
`larger A. If the pressure remains the same, there is thus more force on a larger
`
`vessel.)
`
`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.
`
`-8-
`
`NORRED EXHIBIT 2297 - Page 8
`
`
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