Hi everyone I've bin as usual researching HR and when reading up on invitro wear studies I've seen some figures that I don't really understand here is the extract
. After a median in situ time of 13 months, measurements of the Birmingham Hip Resurfacing femoral heads revealed a median volumetric wear of 2.9 mm3. In one case with a cup abduction angle of 70°, a significantly higher wear of 17.8 mm3 was found. The 2 acetabular components showed volumetric wear of 5.4 and 27.6 mm3 after 14 and 15 months in vivo.
Now those numbers seem high to me but I don't understand the mm3 measuring could someone with a better understanding please explain?
Thanks Danny
Danny-
Can you provide a link to this study?
Cubic millimeters of wear in about a year doesn't seem plausible. Without doing the math it would seem that the acetabular cups would wear out very quickly. And the amount of metal ions in the system that would have to cleared seems very large.
You know that I am going to have to go do some math now to approxiamate the mass of the BHR.
Dan
Hi mate sorry I don't know how to post a link but if you google hip resurfacing wear rates it's one of the first to come up
Danny - use the little URL symbols (insert hyperlink icon) on each side of your URL and it should set it as a hyperlink.
The BHR ace comp is about 4 mm thick, without doing fancy math, I agree measuring volumtric wear in mm3 seems like you'd lose a lot of the mass in short order. Let's assume a common ace comp would be about 56 mm in diameter. With a spherical cup type of mass equation, you ought to be able to figure out the total mass of at least the ace comp. The femoral side might be hard to calculate because the shape is more intricate (on the inside at least).
You might take a look at De Smet's paper on comparing the BHR to the C+ and the wear properties of each. It should be on his website or Pat may have it here on the Surface Hippy.
This could be fun.
Assume that the cup is half of a sphere (it's probably a little less).
the surface area of a sphere of 56 mm diameter is:
4 x pi x r2
The surface area of a half sphere is then
4/2 x 3.14159 x (56mm/2)2
= 4926 mm2
Wear on this surface of 18 mm3 would remove a thickness of
18 mm3/4926 mm2
= 1/274 mm or 0.004 mm
In other words, it would take 27 years to wear off 1/10 of a millimeter.
If the cup is somewhat less than a half sphere then the wear rate would be a little faster, but in the same order of magnitude.
That is some excellent math work. You didn't show all your work though so I will have to give you a B-.
It seemed like a large amount when Danny stated it earlier. But I guess when you spread the wear across the entire surface of a hemisphere it isn't much.
The problems would occur when the wear is concentrated on just part of the surface. Hence the important placement of the device.
Isn't the spacing of the components like 1 mm to start? So after 27 years we could assume the spacing would be 3 mm assuming the same wear on both surfaces. I wonder what this does to the ROM and "looseness" of the joint. Does the joint start to wear more unevenly once the clearance expands?
Lots of neat stuff to explore. I will love it if my implant survives 27 years.
Dan
Quote from: FlbrkMike on September 06, 2011, 03:43:22 PM
Assume that the cup is half of a sphere (it's probably a little less).
the surface area of a sphere of 56 mm diameter is:
4 x pi x r2
The surface area of a half sphere is then
4/2 x 3.14159 x (56mm/2)2
= 4926 mm2
Wear on this surface of 18 mm3 would remove a thickness of
18 mm3/4926 mm2
= 1/274 mm or 0.004 mm
In other words, it would take 27 years to wear off 1/10 of a millimeter.
If the cup is somewhat less than a half sphere then the wear rate would be a little faster, but in the same order of magnitude.
I think its even less than that since the initial wear rate is higher for the first year then it goes into a lower steady state rate.
I did a similar calculation and i think I came up with .6mm/year steady state based on an average of 300000 steps a year. still seems high based on such close tolerances... but its a done deal now. :)
Dan, you make a very good point. If the wear isn't even, and it most likely won't be, the wear will be faster in some areas. This would also result in the shape of the ball and socket being less well matched as time goes on.
And, like Bernie says, it's water under the bridge now for those of us who've already had it done. I hope mine lasts 40 years and that I live to wear it out.
Thanks for the replies guys. I thought it would be a case of me miss interpreting the info. I'll ask Mr Treacy about it tomorrow and see what I can get out of him on the subject.
If when I get mine done I'd be ecstatic if I got 27 years out of it!!!! Here's hoping!
Thank you, guys! I'm glad we have some engineers on this site that are able to figure all this out! If it takes 27 years to wear off 1/10th of a millimeter, I'll be very happy!
^^^ yes that's encouraging isn't it!? I just need to get mine done and all healed up then I can crack on again hopefully!! Having info like this does help to put your mind at ease I find. I love this group!
It's always like that with volumes vs areas, it's always surprising how much volume something can hold, often way more than you expect.
Again, De Smet's paper talks about the wear rates and how wear decreases as the components work into each other. I thought the average cycles per year was more like 1000000. Also, I think the spacing is more like 0.1 mm. There was a recent discussion on spacing, but I can't quickly find it. I think Lop mentioned it somewhere.
Thanks for FlbrMike for the math, that was fun.
I like the math and I assume that with a two hips now done that I can double this up :-) I am very interested to know how the measurements are made - as an engineer it would very interesting to know how they measure the ball diameter - within teh cup to these tight tolerances while still in a person - Cheers - PHill
Phill-
I think they capture the urine from these patients and extract all the metal ions and take their mass. Then the take a blood sample and see how many metal ions are still in the person. Then they add these two values together. ;-)
Dan
Been away for a couple of weeks, but I have to get in on this thread.
Using FlbrkMike's correct formula, I get different results. Using my own device, a 50mm C+ with 170 degree coverage and clearance of 0.1 mil (100 microns)
Four times Pi times Radius squared (using an approximation for the coverage):
Surface area = 4 X 3.14 X 50 X 50 X 170 / 360 square mils
= (about) 15,000 square mils
Assuming an even wear of 3 cubic mils (easier to calculate than 2.9):
Wear depth (per thirteen months) = 3 / 15,000 mils = 0.0002 mils (0.2 microns)
At this wear rate, it would take about two hundred years to wear the device significantly.
So, so long as it is placed correctly, and there is no edge loading, I would not worry.
D.
Lop thanks for that explanation it makes it a little easyier to understand for a maths dummy like me!
Danny, not only get your resurfacing with Mr. Treacy, but also make sure you get all associated information from him, size, angle, clearance, coverage. If it is not useful, it is always interesting.
D.
Thanks mate I'll try and get that info. Your right it's interesting!:-)
Lop, assuming your diameter is 50, you need to divide that by 2 in your calculation to get the radius. That puts your result off by a factor of 4, so you'll have to start worrying in 50 years. I hope you're prepared for that. ;)
Lop-
I am sorry I have to flunk you on your math.
First of all you used the diameter instead of the radius in the formula. So using your femoral head size of 50 a hemisphere would have an area of 2 X 3.1416 X 25 X 25 = 3,927 square mm.
Now you complicated it with wanting to calculate the area of that actual implant which is not a complete hemisphere. You provide the angle of measure as 170 degrees. This means we are 5 degrees short of a hemisphere. The depth of the device then is 25 - sin(5) X 25 = 22.82.
There for the area of your spherical cap is 2 X 3.1416 X 25 X 22.82 = 3,585.
I like your optimism that you will only lose the minimum of 3 cubic mm. I think Mike was thinking even in the worse case of a loss of 18 cubic mm the reduction in thickness is minimal.
You cause me to think about several interesting questions. Are the devices true spherical caps? Is the wear even across the surface? At what increased spacing due to wear does the device become problematic?
Dan
Quote from: FlbrkMike on September 09, 2011, 10:50:20 AM
Lop, assuming your diameter is 50, you need to divide that by 2 in your calculation to get the radius. That puts your result off by a factor of 4, so you'll have to start worrying in 50 years. I hope you're prepared for that. ;)
Aha! Yes Mike, you are right. I faffed that one up didn't I. Oh dear, only get fifty years.
Well I hope as Dr. De Smet says that wear should go down each year. I will be pretty upset if I need revision when I am ninety seven.
Quote from: DGossack on September 09, 2011, 11:08:03 AM
Lop-
I am sorry I have to flunk you on your math.
First of all you used the diameter instead of the radius in the formula. So using your femoral head size of 50 a hemisphere would have an area of 2 X 3.1416 X 25 X 25 = 3,927 square mm.
Now you complicated it with wanting to calculate the area of that actual implant which is not a complete hemisphere. You provide the angle of measure as 170 degrees. This means we are 5 degrees short of a hemisphere. The depth of the device then is 25 - sin(5) X 25 = 22.82.
There for the area of your spherical cap is 2 X 3.1416 X 25 X 22.82 = 3,585.
I like your optimism that you will only lose the minimum of 3 cubic mm. I think Mike was thinking even in the worse case of a loss of 18 cubic mm the reduction in thickness is minimal.
You cause me to think about several interesting questions. Are the devices true spherical caps? Is the wear even across the surface? At what increased spacing due to wear does the device become problematic?
Dan
Yes, yes Dan, I did indeed mess that one up royally by using the diameter.
I did decide to use the fraction 170/360 instead of using sin((180-170)/2) (or whatever) as shallow angles tends towards a ratio anyway.
I most definitely did, however, make the assumption that wear should be even. The two surfaces must remain spherical. If the wear was not even, then they device would quickly fail - as does happen with edge loading.
So realistically it looks promising to last a looooong time then! Hell as I've said before if I get 20 years out of one I'd be mega happy!
Danny-
Thanks for putting it back in perspective. I can get all caught up in the math. At the end of the day it appears that even at the highest levels of wear that they recorded it appears we would get several decades of use. I can live with that.
Dan
Great thread and thanks - makes me feel better and if I get past 100 yr old with these - then old - bits of metal in both hips I will be smiling and thinking of this math - thanks
Dan - Units, units, units!
Also, again I'll reference De Smet's paper on comparison of the BHR and the C+. I think these things are pretty darn spherical. You would get lots more unevenniss if they weren't.
Back the earlier part of the thread, I think a mass balance calculation using urine and blood analysis might not be very accurate due to metals being retained in the system for awhile. Co seems to rid from the body pretty quickly, but Cr, I think is slightly lipophillic and tends to stick around in the body a little longer.
Thanks for the calcs everyone.
Tin-
My comment about taking the mass of metals in the urine and blood was delivered with my tongue firmly planted in my cheek. I am sorry I forgot to use the joking font. :-). This was started by the comment about how they are able to measure the loss in live bodies which of course they can't.
Dan
Dan, I am looking at the icon image of you wearing the big yellow glasses, and I am wondering why your surgeon placed the resurfacing device over your mouth.
Lop-
You obviously haven't been around me much. He used a spare one he had in his office to shut me up! LOL
Dan
Oh, sorry Dan. I didn't get it. Twas lost in the technical world. But I did find out that Co does leave the body pretty quick and Cr stays around a little longer.