My bevel angle calculations

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07-08-2009, 04:23 PM
StraightRazorDave

2 Attachment(s)

My bevel angle calculations

So I was bored last night, so I came up with some formulas for the bevel angle given the thickness of the spine, of the width of the razor, and of the thickness of the tape if it is used.

The first picture is of the derivation of the formulas. To clarify terminology, I used:

x = thickness of the spine
y = width of the razor measured from the edge of the spine to the bevel.
t = thickness of a piece of tape. If 2 or more pieces of tape are used, then just use the combined thickness of the pieces of tape for t.
alpha = angle of the bevel with no tape
beta = additional angle to account for the tape (to get the entire angle we need alpha + beta, hence (2) )

(1) is of the bevel angle using no tape.
(2) is the bevel angle accounting for the thickness of the tape.

The second picture I proved that when t=0, then (2) essentially reduces to (1) since the addition angle (called Beta, the funky B) becomes 0.

I know there have been postings about this in the past, but I thought I would just show how I arrived at my formulas.

Enjoy!
07-08-2009, 05:24 PM
DwarvenChef

Ok to a dyslexic... that looks like alphabet soup...
07-08-2009, 05:58 PM
BeBerlin

Excellent stuff, Dave. Saw this thread: http://straightrazorpalace.com/advan...pers-tape.html ? Feel free to amend and enhance.

Regards,
Robin
07-08-2009, 06:05 PM
Utopian

Quote:

Originally Posted by StraightRazorDave

So I was bored last night, so I came up with some formulas for the bevel angle given the thickness of the spine, of the width of the razor, and of the thickness of the tape if it is used.

It's been a few decades since I needed to use that stuff, and it's apparent that I did not retain it all that well. Anyway, I'd say about a quarter of the people I communicate with about honing ask for an explanation on taping. I'll be sure to provide a link to your post, because I'm sure it will clearly explain everything!
:roflmao
07-08-2009, 06:42 PM
loueedacat

so is tape good or bad?
07-08-2009, 07:00 PM
Ben325e

why not make it simple?

Let:

x = spine width
w= width of razor
n = number of tape layers
t = thickness of tape
Θ = total bevel angle °

total bevel angle:

Θ = 2{arctan[(.5x + nt)/w]}

this accounts for no tape, one or several layers of tape, etc. (when layers of tape = 0 then nt is 0)

For accuracy, in your diagram you have the spine along the x-axis and the blade along the y-axis. This is fine as long as you position the area of spine wear along the x axis such that the spine wear on the left has coordinates (-.5x, 0) and the spine wear on the right has coordinates (.5x, 0), or make some similar compensation if you choose to not have the spine wear directly on the x axis. The width w would then be accurately measured as the distance from the origin to whatever y value the very tip of the blade occupies, assuming a non-warped blade.
07-08-2009, 07:08 PM
StraightRazorDave

Quote:

Originally Posted by Ben325e

why not make it simple?

Let:

x = spine width
w= width of razor
n = number of tape layers
t = thickness of tape
Θ = total bevel angle

total bevel angle:

Θ = 2{arctan[(.5x + nt)/w]}

this accounts for no tape, one or several layers of tape, etc. (when layers of tape = 0 then nt is 0)

For accuracy, in your diagram you have the spine along the x-axis and the blade along the y-axis. This is fine as long as you position the area of spine wear along the x axis such that the spine wear on the left has coordinates (-.5x, 0) and the spine wear on the right has coordinates (.5x, 0), or make some similar compensation if you choose to not have the spine wear directly on the x axis. The width w would then be accurately measured as the distance from the origin to whatever y value the very tip of the blade occupies, assuming a non-warped blade.

Ah, I like it. Thanks. I didn't measure the width of the blade directly, but used a diagonal measurement from the bevel to the edge of the spine. I figured this would be easier to measure physically then the actualy width, since you would need to measure the cross-section essentially, which would be impossible if you were measuring the bevel angle in the middle on the blade. Does that make sense? In my diagram, the actual width of the razor is z, but if you actually measure your razor it would be easier to measure y, the hypotenuse.

I like your idea of using an integral multiple of the number of tape layers. I just figured you could just substitute the total thickness in for t, but your way is more eligant.
07-08-2009, 07:19 PM
BeBerlin

Honing_Angles_Matrix.xls?
07-08-2009, 07:20 PM
Ben325e

Quote:

Originally Posted by StraightRazorDave

Ah, I like it. Thanks. I didn't measure the width of the blade directly, but used a diagonal measurement from the bevel to the edge of the spine. I figured this would be easier to measure physically then the actualy width, since you would need to measure the cross-section essentially, which would be impossible if you were measuring the bevel angle in the middle on the blade. Does that make sense? In my diagram, the actual width of the razor is z, but if you actually measure your razor it would be easier to measure y, the hypotenuse.

I like your idea of using an integral multiple of the number of tape layers. I just figured you could just substitute the total thickness in for t, but your way is more eligant.

I can get on board with that - the measurement would be easier. So:

Let:

x = spine width
y= distance from spine/hone contact point to tip of bevel
n = number of tape layers
t = thickness of tape
Θ = total bevel angle °

total bevel angle:

Θ = 2{arcsin[(.5x + nt)/y]}
07-08-2009, 07:32 PM
StraightRazorDave

Quote:

Originally Posted by Ben325e

I can get on board with that - the measurement would be easier. So:

Let:

x = spine width
y= distance from spine/hone contact point to tip of bevel
n = number of tape layers
t = thickness of tape
Θ = total bevel angle °

total bevel angle:

Θ = 2{arcsin[(.5x + nt)/y]}

I like it! But your new y would have to be measured again if you added tape, since it would be larger if you added tape. But it makes complete sense to me. :tu
07-08-2009, 07:40 PM
Ben325e

Quote:

Originally Posted by BeBerlin

Honing_Angles_Matrix.xls?

Robin, what type of tape are you using for these calculations? On the spreadsheet it shows that the thickness of the tape is just under a third of a centimeter, and that's some ultra thick tape. That's just under an eight of an inch for one layer, on one side.

I'd suggest using the measurements for 3M Super 33+ electrical tape. It's supposed to be 7 mil thick, but I just put on my digital calipers and I get .18 mm on average. That's 0.18 cm vs 3.0.

3M Super 33+ is quality electrical tape readily available to most on the forums. If we are going to use something as a 'standard measurement', then it should be good quality stuff, I'd think.
07-08-2009, 07:50 PM
BeBerlin

Quote:

Originally Posted by Ben325e

Robin, what type of tape are you using for these calculations? On the spreadsheet it shows that the thickness of the tape is just under a third of a centimeter, and that's some ultra thick tape. That's just under an eight of an inch for one layer, on one side.

My bad, I thought I'd updated the file in the meantime. It's 0.3mm, not 0.3cm. Regular electrical tape.

Quote:

Originally Posted by Ben325e

3M Super 33+ is quality electrical tape readily available to most on the forums. If we are going to use something as a 'standard measurement', then it should be good quality stuff, I'd think.

Absolutely, which is why I am using tesa Isolierband. :popcorn:

Doesn't really matter, though. Just fill in the thickness of your favourite tape in the spreadsheet, and measuring happiness will ensue.
07-08-2009, 08:05 PM
JimmyHAD

Quote:

Originally Posted by Ben325e

3M Super 33+ is quality electrical tape readily available to most on the forums. If we are going to use something as a 'standard measurement', then it should be good quality stuff, I'd think.

That is what I use. Made in Minnesota. For those of use in the USA, buy American. The job you save may be your own.
07-08-2009, 08:19 PM
fccexpert

Okay......
07-08-2009, 08:27 PM
Ben325e

Quote:

Originally Posted by StraightRazorDave

I like it! But your new y would have to be measured again if you added tape, since it would be larger if you added tape. But it makes complete sense to me. :tu

x = spine width
y = distance from spine/hone contact point to tip of bevel with no tape on spine
n = number of tape layers
t = thickness of tape
Θ = total bevel angle °

total bevel angle:

Θ = 2{arctan[(.5x + nt)/{ycos(arcsin{.5x/y})}]}

arcsin(.5x/y) gets you the angle on one side of the bevel, then y times cosine of the resultant angle gets you the width of the razor that I was formerly calling w.

This allows for w to remain constant no matter how many layers of tape are added, eliminating the need to 1.) remeasure the hypotenuse every time you add a layer of tape, and 2.) measure w initially, which as you pointed out could be difficult.

Note that I newly defined y. Defining w like this is definitely losing some of the elegance! :)
07-08-2009, 08:57 PM
Bill S

Quote:

Originally Posted by JimmyHAD

That is what I use. Made in Minnesota. For those of use in the USA, buy American. The job you save may be your own.

Quote:

Originally Posted by fccexpert

Okay......

These are the only two posts that I read in this thread.

As a Polymath I found the others, well...you know.:D
07-08-2009, 09:09 PM
gratewhitehuntr

wtf is Polymath?

and since when does it take a whole sheet of paper to find an angle based on the height and base width of a triangle ??

I must be missing something here (am I Polymath?)

Of course my Pythagorean theorem involves making a square with another identical triangle, then calculating IT'S area, then dividing by two.

what is it I am missing here?
07-08-2009, 09:41 PM
Ben325e

Quote:

Originally Posted by gratewhitehuntr

wtf is Polymath?

http://en.wikipedia.org/wiki/Polymath

and since when does it take a whole sheet of paper to find an angle based on the height and base width of a triangle ??

It doesn't. Dave was just being thorough in showing his calculations.

I must be missing something here (am I Polymath?)

The only thing missing is that we aren't finding an angle based on the height and base width of a triangle. We made one equation that gives the total angle on a straight razor whether you hone it with no tape or any reasonable number of layers of tape. Did you use three layers of tape? The equation works. Did you use no tape? it works. One layer? works.

Of course my Pythagorean theorem involves making a square with another identical triangle, then calculating IT'S area, then dividing by two.

That's the equation for the area of a triangle. 1/2 base times height.

what is it I am missing here?

Pythagorean theorem is a² + b² = c². It finds the length of the hypotenuse.

I hope this doesn't come across sarcastic - I'm just trying to answer the questions. :)
07-08-2009, 09:52 PM
kevint

none of my razors have a straight spine thickness, they all have distal taper thicker at the toe, one even has a narrower width at the heel , at least 1/8" wider at the toe-(my fav, and it was made that way)

whattya do then?
07-08-2009, 09:57 PM
hoglahoo

Quote:

Originally Posted by kevint

whattya do then?

fix it with tape?
07-08-2009, 09:59 PM
StraightRazorDave

Quote:

Originally Posted by kevint

none of my razors have a straight spine thickness, they all have distal taper thicker at the toe, one even has a narrower width at the heel , at least 1/8" wider at the toe-(my fav, and it was made that way)

whattya do then?

Well the formulas (either mine or Ben's) work for only a section of the blade. So you could measure the thickness of the spine and the distance from the spine to the edge at different points along the edge and get different angles. For non-straight razors (i.e. the edge isn't straight), it would come to no suprise to me that the angle changes somewhat as you progress from the heel to the toe. The angle doesn't have to be constant throughout the edge. I think only a perfectly straight edge with a perfectly uniform wear on the edge and the spine will produce the exact same bevel angle throughout the edge.

Well...I guess you could have a smiling razor that had the same angle throughout given that the distance between the edge and the spine was constant throughout the entire length of the razor (i.e. they both have the exact same curvature), as well as having perfectly even wear on the edge and the spine.
07-08-2009, 10:06 PM
kevint

Thanks Dave that makes sense.

But you know; what I meant was, how does one get that bored?
07-08-2009, 10:07 PM
Ben325e

Quote:

Originally Posted by kevint

none of my razors have a straight spine thickness, they all have distal taper thicker at the toe, one even has a narrower width at the heel , at least 1/8" wider at the toe-(my fav, and it was made that way)

whattya do then?

It depends. If the width of the razor (from spine to edge) varies proportionally to the spine thickness, then nothing. The formula stays the same.

If the razor width remains the same but the spine thickness varies down the length of the razor, then you can take several measurements, apply the formula for each set of measurements, then interpolate the data for 'close enough' measurements.

Otherwise (if you want just one equation), you can set up an integral (calculus II techniques). But, since the variation in thickness is probably pretty constant per unit of length, I'd just take several measurements and then plug the numbers in the formula.
07-08-2009, 10:15 PM
StraightRazorDave

Quote:

Originally Posted by kevint

Thanks Dave that makes sense.

But you know; what I meant was, how does one get that bored?

I have a math degree, so what entertains me might be different than most people. :rofl2:

Quote:

Originally Posted by Ben325e

Otherwise (if you want just one equation), you can set up an integral (calculus II techniques). But, since the variation in thickness is probably pretty constant per unit of length, I'd just take several measurements and then plug the numbers in the formula.

It would be AWESOME if you could use calculus in calculating the angles. Would the resulting answer not be a value but a function dependent on one variable, how far along the razor you'd like to know the angle of? That would be cool. :D
07-08-2009, 11:10 PM
gratewhitehuntr

Quote:

Originally Posted by Ben325e

I hope this doesn't come across sarcastic - I'm just trying to answer the questions. :)

not at all

just didn't understand the equation (of course I didn't really look, like it would matter anyway)

and I wasn't trying to say anything about the PT applying here but rather to give an example of my non-use of formulas
as I explained to my high school math teacher when he flunked me for not showing my work (the work he wanted to see done HIS way)
"The idea behind swinging a hammer is to sink a nail, not to exercise my damn arm!!"

I hope all this work turns into something (like an online calculator LOL)
07-09-2009, 12:04 AM
AFDavis11

Quote:

Originally Posted by Ben325e

I hope this doesn't come across sarcastic - I'm just trying to answer the questions. :)

You kidding? Last month we were shaving whiskers with vibratory harmonics . . .
07-09-2009, 05:28 AM
BeBerlin

1 Attachment(s)

Quote:

Originally Posted by StraightRazorDave

I have a math degree, so what entertains me might be different than most people. :rofl2:

Oh, lovely. I have always hated maths. Would you kindly help a heathen find the flaw in the following?

The picture below is supposed to show the the profile of a razor. The red line shows the triangular geometry of the razor and the enclosed bevel angle. The razor is divided in 2 identical rectangular triangles.

Calculating the acute angle for any given rectangular triangle - as far as I remember and understand it - is: Sin a = The half of B divided by A. We would be using the half of B, and we're only calculating half of the razor's bevel angle. However, we're not looking for Sin a, but a itself:a = arcsin ((B/2)/A). And apparently, that's only half of the razor's bevel angle. That's why in the spreadsheet the complete angle = 2*(arcsin ((B/2)/A).

Now, when we tape the spine as shown by the black line in the drawing, and we assume that the thickness of typical electrical insulation tape is 0.15mm (keeping in mind that one layer thickens the spine twice, i.e. once on each side, as shown in the image), we will end up with a "measured spine after taping" which will be 2*0.15=0.3mm thicker than before. This brings the formula to: 2*(arcsin ((B+(2*tape_tickness)/2)/A).

Right?

By the way, I should think that if you need to know the exact thickness of your tape, stick 10 layers together, measure them and divide by 10 (this also divides the inaccuracy of the measurement by 10).

Attachment 27405

Quote:

Originally Posted by StraightRazorDave

It would be AWESOME if you could use calculus in calculating the angles. Would the resulting answer not be a value but a function dependent on one variable, how far along the razor you'd like to know the angle of? That would be cool. :D

That would be most cool.

Thanks!
Robin
07-09-2009, 05:48 AM
StraightRazorDave

Quote:

Originally Posted by BeBerlin

...
That's why in the spreadsheet the complete angle = 2*(arcsin ((B/2)/A).
...
...we will end up with a "measured spine after taping" which will be 2*0.15=0.3mm thicker than before. This brings the formula to: 2*(arcsin ((B+(2*tape_tickness)/2)/A).

The first equation is correct, but there is a small flaw in the second equation. You are correct in saying the width of the spine changes giving you the extra term to be added to B, but it shouldn't be 2*tape_thickness since you are only considering half the angle in this triangle, thus only 1*tape_thickness, the thickness of the other half of the tape is already accounted for in the 2* in the beginning of the equation before the arcsin term. Also, A in the second equation is not the same A as in the first one. When you add tape, you are not only increasing the thickness of the spine, but also the distance (the hypotenuse "A") from the edge of the bevel to the spine.

So you could the formula: 2*(arcsin ((B+tape_tickness/2)/A), but the A in this formula would have to be the measured distance from the spine w/tape to the bevel. Note that I took out the 2* coefficient in front of the tape_thickness variable.

If you look at some of the previous posts between Ben and I, we were discussing this problem. But, it was then accounted for. :)

I hope what I wrote makes sense. I can always write up what I mean by hand and scan it if I need to be more clear. It's almost 2 AM, so it may not be the clearest.

P.S. I like to write the symbol for inverse sin and tan, basically just a little superscript -1, but it's same as writing arcsin or arctan. It's just a personal preferrence for some reason....Helps me remember that it's essentially an inverse function. :)
07-09-2009, 12:37 PM
BeBerlin

Quote:

Originally Posted by StraightRazorDave

The first equation is correct, but there is a small flaw in the second equation. You are correct in saying the width of the spine changes giving you the extra term to be added to B, but it shouldn't be 2*tape_thickness since you are only considering half the angle in this triangle, thus only 1*tape_thickness, the thickness of the other half of the tape is already accounted for in the 2* in the beginning of the equation before the arcsin term. Also, A in the second equation is not the same A as in the first one. When you add tape, you are not only increasing the thickness of the spine, but also the distance (the hypotenuse "A") from the edge of the bevel to the spine.

Big fat warning: I hate maths. Always have, always will. It's a truly mutual relationship, by the way.

That said, I am not entirely certain that what you said is entirely accurate. Take a look at the drawing again. We measure B, which is the total spine thickness. Suppose we did not calculate the taped bevel, but just measure the taped spine, in the same fashion as we measure the un-taped spine. Obviously, we are then measuring 2 extra layers (I tried to illustrate that in the drawing as well). So when using the same formula, in the taped version the spine thickness is B+2 layers of tape. It is true that you can easily rewrite the formula (B+2T)/2 = B/2+T. Both are correct. I believe my variant stays closer to what would happen if you were to measure the taped spine instead of calculating it. Quite honestly, I thought that would be simpler.

The rest is, I'm afraid, slightly beyond me. I think that by widening B, also A will become very marginally longer. However, a spine is usually rounded and neither does tape create a sharp corner where it is folded over the rims of the spine. I couldn't think of a formula that completely correlates with how exactly the tape folds around the rim. Moreover, I think this factor will only alter the outcome in amounts far beyond the precision of our measurements (using the Verniers calipers) and far beyond the normal variations in spine thickness and blade width that can be found along the same blade. Variations in thickness of the tape itself will nullify the intended extra precision.

I'm sure this will be an interesting exercise for you mathematicians (and, of course, the Polymaths), but for practical use in real life on real razors, it seems a bit over the top. Personally, knowing the bevel angle within half a degree is more than accurate for my purposes. ;)

Quote:

Originally Posted by StraightRazorDave

I hope what I wrote makes sense. I can always write up what I mean by hand and scan it if I need to be more clear. It's almost 2 AM, so it may not be the clearest.

I think it would be fun to have that in the Excel spreadsheet for practical purposes.

Thanks a lot!
Robin
07-09-2009, 01:24 PM
Bruce

This is all too fascinating!! I have to admit some confusion about the angle in discussion. Have we come to any agreement on how we are describing the angle as it pertains to half the thickness of the spine or are we talking about the entire included angle which is described by both bevels meeting at one edge?

This may be important as we are told that we need to hold the razor at a 30 degree angle to our properly stretched faces. Now is one side of that 30 degree triangle taken down the center of the the razor or is that side determined by the far side of the edge and the far corner of the spine, as it relates to the face, or is it the near side of the edge and the near corner of the spine as it relates to the face? And has anyone taken into account the thickness of the lather?

And does arcsin mean really, really sharp or is that function yupswell? I have to admit to getting a bit sloppy with my angles, I know there is no excuse for that, but in my defense I am really fussy with yupswells.

Carry on,
07-09-2009, 02:45 PM
hoglahoo

Bruce, I've never quite grasped the 30 degree shaving angle suggestion either. If the bevel angle is 20 degrees, does that mean I should lift the razor off my face another 10 degrees, another 20 degrees, or another 30 degrees? Even if I knew, I don't think it would help as I really cannot afford to attempt taking such a measurement while shaving :shrug:
07-13-2009, 05:00 AM
StraightRazorDave

Quote:

Originally Posted by BeBerlin

Big fat warning: I hate maths. Always have, always will. It's a truly mutual relationship, by the way.

That said, I am not entirely certain that what you said is entirely accurate. Take a look at the drawing again. We measure B, which is the total spine thickness. Suppose we did not calculate the taped bevel, but just measure the taped spine, in the same fashion as we measure the un-taped spine. Obviously, we are then measuring 2 extra layers (I tried to illustrate that in the drawing as well). So when using the same formula, in the taped version the spine thickness is B+2 layers of tape. It is true that you can easily rewrite the formula (B+2T)/2 = B/2+T. Both are correct. I believe my variant stays closer to what would happen if you were to measure the taped spine instead of calculating it. Quite honestly, I thought that would be simpler.

The rest is, I'm afraid, slightly beyond me. I think that by widening B, also A will become very marginally longer. However, a spine is usually rounded and neither does tape create a sharp corner where it is folded over the rims of the spine. I couldn't think of a formula that completely correlates with how exactly the tape folds around the rim. Moreover, I think this factor will only alter the outcome in amounts far beyond the precision of our measurements (using the Verniers calipers) and far beyond the normal variations in spine thickness and blade width that can be found along the same blade. Variations in thickness of the tape itself will nullify the intended extra precision.

I'm sure this will be an interesting exercise for you mathematicians (and, of course, the Polymaths), but for practical use in real life on real razors, it seems a bit over the top. Personally, knowing the bevel angle within half a degree is more than accurate for my purposes. ;)

I think it would be fun to have that in the Excel spreadsheet for practical purposes.

Thanks a lot!
Robin

Sorry for the late response, haven't checked this thread in a bit! What you mentioned in the first long paragraph could work just fine too. You can easily just measure the width of the spine and the bevel with the tape on and substitute these values into the original (simpler) equation and not have the need for an extra variable (i.e. tape thickness) and you wouldn't even need to know the thickness of the tape in the first place. This is, like you mentioned, simpler and more practical. To be honest I derived the second formula just to see what it would look like and for fun. But it would be more prone to error since you are using a given value of the thickness of the tape when it may not be as exact as if you actually measured the thickness of the tape on the razor. Also, if you added tape you'd be removing some metal from the edge also, shortening A as labelled in your diagram. I'm sure you could find some fancy-dancy (yes, that is a mathematical term) formula that measures the bevel angle but also takes into account the amount of the wear on the edge given the grit used, the weight of the honer, the day of the month, what colour shirt you are wearing.......etc.....etc....

But for all everyday bevel angle calculations (which I honesly don't do or really care about), it would surely be easier to make the measurements with the tape actually on the razor and use a simple equation. But what's the fun in being practical when you can use some convoluted math equation that makes everything more difficult? :rofl2:
07-15-2009, 01:47 AM
SlickWillyTFCF

So judging by this thread: since I'm a newbie and don't have a degree in mathmatics, I should wrap my razor in my favorite, high-quality electrical tape and send it to my local scientist to have it honed? :)
07-15-2009, 01:58 AM
JMS

I read through this whole damned thread to find a gssixgun post for pure entertainment...and the bastard didn't even post!:rant:

You owe me 5 minutes of my life Glen!!...Of course you can always buy me a beer instead.:beer1:
07-15-2009, 03:59 AM
gssixgun

Quote:

Originally Posted by JMS

I read through this whole damned thread to find a gssixgun post for pure entertainment...and the bastard didn't even post!:rant:

You owe me 5 minutes of my life Glen!!...Of course you can always buy me a beer instead.:beer1:

LMFAO Mark...
I am waiting for the conclusion to the thread, which from my real life experience, I will guess that each layer of 3M tape will add just barely over 1/2 of one degree of angle change to the average edge, which will accomplish much to save the spine and the stone but little else that is provable...