Regular Member GratefulHead Posted December 20, 2003 Regular Member Share Posted December 20, 2003 Since I was interested in quantifying the improvement from my hair transplant (and I'm an engineer by trade), I attempted to calculate my post-op hair density increase. I'm sharing my methods here for those of you who are technically minded and would like to do the same. One disclaimer right up front is that these are very rough approximate calculations! More exact calculations would require additional scalp/hair measurements, a little gradient calculus, and more computation time on my part. The rough calculations presented here are good enough for my purposes and are vastly more meaningful than just qualitatively saying "my hair looks better after my transplant than before". It also will give those who recently underwent a transplant something quantitative to look forward to (and to keep busy with ) while they are in the long multi-month waiting period. Lastly, it should help people set realistic expectations for a given graft count transplant. Note: These calculations were actually performed with 12+ significant digits of precision carried all the way through, but all results (both intermediate and final) shown below are rounded off for simplicity/readability. This explains why you will obtain slightly different answers if you attempt to check my calculations using the intermediate rounded-off values. Without further ado, on to the calculations... AREA OF THINNING/BALD SCALP: ---------------------------- Since I'm approaching a Norwood 5A, I have an elliptical (that is, oval) thinning zone incorporating my front, midscalp, and vertex (a.k.a. crown). The length of my thin area "major" and "minor" ellipse diameters, d1 and d2, are 16cm and 10cm, respectively. Since a radius ® is defined as half a diameter, r1 = 8cm and r2 = 5cm. Finally, the area of my thinning ellipse (approximating to be planar) is r1 x r2 x Pi = (8cm)(5cm)(3.14159) = 125.7cm^2 (or centimeters squared). AREA OF DONOR STRIP: -------------------- My donor strip was an approximate 26cm x 1.2cm hemi-ellipse. A hemi-ellipse is like an oval with two sides somewhat flattened straight. Given the available measurement data and based on Dr. True's estimate, I will assume the middle 22cm of the hemi-elliptical donor strip is a rectangle and the last 2cm on each end are both "half-ellipses" (in actuality the strip is curved a little like a banana). Hence, the center rectangle of the donor strip is 22cm x 1.2cm = 26.4cm^2. Furthermore, putting both of the end 2cm half-ellipses together makes roughly a whole ellipse (or oval) with a major diameter of d1 = 4cm and a minor diameter d2 = 1.2cm. Thus, from before radius r1 = 2cm and r2 = 0.6cm. Then, the area at the ends of the donor strip is about (2cm)(0.6cm)(3.14159) = 3.77cm^2. Next, the total donor strip area is sum of the center rectangle and end half-ellipses or 26.4cm^2 + 3.77cm^2 = 30.17cm^2. To complicate matters even further, the donor region is stretched around 10% - 20% by injecting fluid to lift the skin away from the skull before excising it out. So finally we can conclude the original unstretched donor area is say 80% of the removed donor strip just computed or (30.17cm^2)(0.8) = 24.14cm^2. DONOR DENSITY: -------------- Of my 1503 follicular-unit grafts, 765 were single-hairs (51%), 541 were double-hairs (36%), and 197 were triples or rarely quadruples (13%). This adds up to 2458 total hairs (assuming 20 quadruples) and averages to 1.64 hairs/follicular unit graft. Since a typical young person has around 2.1 hairs/follicular unit, I am 22% below average density. The unstretched donor density is 1503 grafts / 24.14cm^2 = 62.27 grafts/cm^2 (verses a typical 80 - 120 grafts/cm^2 you are born with). However, people with fairly coarse hair like myself typically have lower density. The waviness of my hair is another offsetting factor. Lastly, 62.27 grafts/cm^2 x 1.64 hairs/graft = 101.84 hairs/cm^2 on average in the original unstretched donor area. Consistent with the above numbers, a typical young person has 168 - 252 hairs/cm^2. RECIPIENT DENSITY: ------------------ If I apply my average donor density over the entire recipient area, we have 101.84 hairs/cm^2 x 125.7cm^2 = 12,798 hairs originally before thinning. If I estimate my overall progressive thinning left 30% of my original hair on average, we obtain 12,798 hairs x 0.3 = 3839 hairs before my first transplant. If we assume a 98% graft survival rate (my personal estimate), we see 2458 transplanted hairs x 0.98 = 2409 net hairs are added to the recipient area. 2409 transplanted hairs / 3839 "before" hairs = 62.7% increase. (2409 net hairs added + 3839 "before" hairs) / 12,798 "original" hairs = 48.8% of original hair is now present after first transplant. Since a typical rule of thumb is hair loss isn't very noticeable until the density drops by 50%, I shouldn't be too far from appearing follicularly full after the transplant completely grows in. This may be deceiving, however, when one considers that my original thinning was asymmetrical (uneven) and my transplant was heavily weighted towards the front (around 25 grafts/cm^2) and thinner areas as opposed to the midscalp/vertex and higher density areas. From a high level, approximately 800 grafts were placed in the front and around 700 in the midscalp/vertex area. Hence, I will surely have a thinner midscalp/vertex area and a thicker front region and will likely require another transplant to improve density in some places. Dr. True told me "further treatment (would be) needed to add density..." at my consultation. At 6.5 months out from my transplant, I do indeed have full-coverage, but with the expected thin look on my midscalp and vertex. Aesthetically, I would say my hair appearance agrees with these calculations and looks 2/3'rds better (without camouflage) than it did pre-transplant. With the aid of Toppik, I have the appearance of pretty much full normal density. This is SOOO AWESOME to me because I haven't had that look for a good 7 years!! I invite you to extend and improve the calculation techniques presented here by employing densitometry measurements, accounting for density gradients, approximating 3-dimensional recipient/donor surface areas, and any other accuracy enhancement you feel so inclined to pursue. Some doctors would probably do well by improving and automating these calculations in a spreadsheet (i.e., Excel) and offering these "rough estimates" to their clients with appropriate unpredictability disclaimer statements. I hope this append is understandable and will be useful to some of you for planning purposes. GratefulHead [This message was edited by GratefulHead on December 19, 2003 at 08:10 PM.] Link to comment Share on other sites More sharing options...

Regular Member GratefulHead Posted December 20, 2003 Author Regular Member Share Posted December 20, 2003 Since I was interested in quantifying the improvement from my hair transplant (and I'm an engineer by trade), I attempted to calculate my post-op hair density increase. I'm sharing my methods here for those of you who are technically minded and would like to do the same. One disclaimer right up front is that these are very rough approximate calculations! More exact calculations would require additional scalp/hair measurements, a little gradient calculus, and more computation time on my part. The rough calculations presented here are good enough for my purposes and are vastly more meaningful than just qualitatively saying "my hair looks better after my transplant than before". It also will give those who recently underwent a transplant something quantitative to look forward to (and to keep busy with ) while they are in the long multi-month waiting period. Lastly, it should help people set realistic expectations for a given graft count transplant. Note: These calculations were actually performed with 12+ significant digits of precision carried all the way through, but all results (both intermediate and final) shown below are rounded off for simplicity/readability. This explains why you will obtain slightly different answers if you attempt to check my calculations using the intermediate rounded-off values. Without further ado, on to the calculations... AREA OF THINNING/BALD SCALP: ---------------------------- Since I'm approaching a Norwood 5A, I have an elliptical (that is, oval) thinning zone incorporating my front, midscalp, and vertex (a.k.a. crown). The length of my thin area "major" and "minor" ellipse diameters, d1 and d2, are 16cm and 10cm, respectively. Since a radius ® is defined as half a diameter, r1 = 8cm and r2 = 5cm. Finally, the area of my thinning ellipse (approximating to be planar) is r1 x r2 x Pi = (8cm)(5cm)(3.14159) = 125.7cm^2 (or centimeters squared). AREA OF DONOR STRIP: -------------------- My donor strip was an approximate 26cm x 1.2cm hemi-ellipse. A hemi-ellipse is like an oval with two sides somewhat flattened straight. Given the available measurement data and based on Dr. True's estimate, I will assume the middle 22cm of the hemi-elliptical donor strip is a rectangle and the last 2cm on each end are both "half-ellipses" (in actuality the strip is curved a little like a banana). Hence, the center rectangle of the donor strip is 22cm x 1.2cm = 26.4cm^2. Furthermore, putting both of the end 2cm half-ellipses together makes roughly a whole ellipse (or oval) with a major diameter of d1 = 4cm and a minor diameter d2 = 1.2cm. Thus, from before radius r1 = 2cm and r2 = 0.6cm. Then, the area at the ends of the donor strip is about (2cm)(0.6cm)(3.14159) = 3.77cm^2. Next, the total donor strip area is sum of the center rectangle and end half-ellipses or 26.4cm^2 + 3.77cm^2 = 30.17cm^2. To complicate matters even further, the donor region is stretched around 10% - 20% by injecting fluid to lift the skin away from the skull before excising it out. So finally we can conclude the original unstretched donor area is say 80% of the removed donor strip just computed or (30.17cm^2)(0.8) = 24.14cm^2. DONOR DENSITY: -------------- Of my 1503 follicular-unit grafts, 765 were single-hairs (51%), 541 were double-hairs (36%), and 197 were triples or rarely quadruples (13%). This adds up to 2458 total hairs (assuming 20 quadruples) and averages to 1.64 hairs/follicular unit graft. Since a typical young person has around 2.1 hairs/follicular unit, I am 22% below average density. The unstretched donor density is 1503 grafts / 24.14cm^2 = 62.27 grafts/cm^2 (verses a typical 80 - 120 grafts/cm^2 you are born with). However, people with fairly coarse hair like myself typically have lower density. The waviness of my hair is another offsetting factor. Lastly, 62.27 grafts/cm^2 x 1.64 hairs/graft = 101.84 hairs/cm^2 on average in the original unstretched donor area. Consistent with the above numbers, a typical young person has 168 - 252 hairs/cm^2. RECIPIENT DENSITY: ------------------ If I apply my average donor density over the entire recipient area, we have 101.84 hairs/cm^2 x 125.7cm^2 = 12,798 hairs originally before thinning. If I estimate my overall progressive thinning left 30% of my original hair on average, we obtain 12,798 hairs x 0.3 = 3839 hairs before my first transplant. If we assume a 98% graft survival rate (my personal estimate), we see 2458 transplanted hairs x 0.98 = 2409 net hairs are added to the recipient area. 2409 transplanted hairs / 3839 "before" hairs = 62.7% increase. (2409 net hairs added + 3839 "before" hairs) / 12,798 "original" hairs = 48.8% of original hair is now present after first transplant. Since a typical rule of thumb is hair loss isn't very noticeable until the density drops by 50%, I shouldn't be too far from appearing follicularly full after the transplant completely grows in. This may be deceiving, however, when one considers that my original thinning was asymmetrical (uneven) and my transplant was heavily weighted towards the front (around 25 grafts/cm^2) and thinner areas as opposed to the midscalp/vertex and higher density areas. From a high level, approximately 800 grafts were placed in the front and around 700 in the midscalp/vertex area. Hence, I will surely have a thinner midscalp/vertex area and a thicker front region and will likely require another transplant to improve density in some places. Dr. True told me "further treatment (would be) needed to add density..." at my consultation. At 6.5 months out from my transplant, I do indeed have full-coverage, but with the expected thin look on my midscalp and vertex. Aesthetically, I would say my hair appearance agrees with these calculations and looks 2/3'rds better (without camouflage) than it did pre-transplant. With the aid of Toppik, I have the appearance of pretty much full normal density. This is SOOO AWESOME to me because I haven't had that look for a good 7 years!! I invite you to extend and improve the calculation techniques presented here by employing densitometry measurements, accounting for density gradients, approximating 3-dimensional recipient/donor surface areas, and any other accuracy enhancement you feel so inclined to pursue. Some doctors would probably do well by improving and automating these calculations in a spreadsheet (i.e., Excel) and offering these "rough estimates" to their clients with appropriate unpredictability disclaimer statements. I hope this append is understandable and will be useful to some of you for planning purposes. GratefulHead [This message was edited by GratefulHead on December 19, 2003 at 08:10 PM.] Link to comment Share on other sites More sharing options...

Senior Member EastCoast Posted February 14, 2004 Senior Member Share Posted February 14, 2004 Have a question. Ht looks natural when hairs are about a mill. apart. How many hairs does it take for a nw 6 to be covered with hairs 1 mill. apart. Link to comment Share on other sites More sharing options...

Regular Member GratefulHead Posted February 20, 2004 Author Regular Member Share Posted February 20, 2004 Hi Eastcoast, The answer to your question depends on exactly how big your bald Norwood 6 area is. A Norwood 6 has an elliptical (that is, oval) thinning zone incorporating the front, midscalp, and vertex (a.k.a. crown) areas. The length of the "major" and "minor" bald ellipse (or oval) diameters, d1 and d2, should be measured. Since a radius ® is defined as half a diameter, r1 = d1/2 and r2 = d2/2. Finally, the area ("A") of the bald ellipse (approximating to be planar) is r1 x r2 x Pi = (r1)(r2)(3.14159) = "A" cm^2 (or centimeters squared). Your stated goal of 1 hair every millimeter equates to 100 hairs/cm^2. Hence, multiply the above bald area "A" by 100 to obtain the approximate number of hairs necessary to look follicularly normal. I hope this answers your question. Rugger, Thank you for acting as a positive employment reference! Maybe we can discover hair on Mars??? GratefulHead Link to comment Share on other sites More sharing options...

Regular Member Bushy (fka jontew) Posted March 13, 2004 Regular Member Share Posted March 13, 2004 I like cottonballs. In addition to being a patient of Dr. Hasson, I act as the legal counsel for Hasson and Wong. Link to comment Share on other sites More sharing options...

## Recommended Posts

## Create an account or sign in to comment

You need to be a member in order to leave a comment

## Create an account

Sign up for a new account in our community. It's easy!

Register a new account## Sign in

Already have an account? Sign in here.

Sign In Now