Thread: Latest custom builds

Originally Posted by silverloaf

nope. its roughly 30 degrees effective cutting angle. good grief, thats about what my axes are, for some reason I pictured microtomes being a more acute angle than that.
I might be wrong on that angle, I don't know formulas too well so I had to go oldschool. now the bigger question is this: what is the actual width of the "spine" created using that frameback? the diameter of the frame is 16mm but the points where it contacts the hone are not across the diameter of the frame but slightly down from the "equator." so the measure between those points would be the spine width...... what is it?

30 mm x 16mm. Distance where the frameback spine hits the hone and to the edge diameter of the frameback at the point of contacting the hone! Seems you have not divided by 2 somewhere! Lord knows! I certainly don't!

30mm from the point of spine on hone to cutting edge. diameter of frame is 16mm, radius of 8mm. a right triangle using the 30mm and 8mm still only gives you half the effective cutting angle. so the effective angle is roughly 30 unless im really missing something. you wouldn't divide by 2 to get effective angle. im gonna feel incredibly stupid if im this wrong after drawing out the stinkin thing! hahaha

I dunno. Seems some table comes into play. Simpletons like us cannot comprehend!
My fine friend Tarkus wil explain! Tomorrow!

Originally Posted by sharptonn

30 mm x 16mm. Distance where the frameback spine hits the hone and to the edge diameter of the frameback at the point of contacting the hone! Seems you have not divided by 2 somewhere! Lord knows! I certainly don't!

I aint seein' it man, I may have to rescind my "nope!" im reading it as the frame is tubular and has a diameter of 16mm. is it 16mm across the point where it hits the hone? then im reading that the height fro cutting edge to where the frame hits hone is 30mm. I don't think im confused anywhere so far???? so if I made a right triangle with 30mm being the long side and 8mm being the short then my cutting angle is 14.9ish. I would double that to get my effective cutting angle since that gives me only one side of my blade??????????

It is about 30 degree angle, and the effect from the hone not touching directly at the diameter is second order about as significant as the effect from the exact position of the cylindrical shim on top of the blade. Overall thats a tolerance of about 2 degrees, so 30 plus/minus 2

Originally Posted by silverloaf

nope. its roughly 30 degrees effective cutting angle. good grief, thats about what my axes are, for some reason I pictured microtomes being a more acute angle than that.
I might be wrong on that angle, I don't know formulas too well so I had to go oldschool. now the bigger question is this: what is the actual width of the "spine" created using that frameback? the diameter of the frame is 16mm but the points where it contacts the hone are not across the diameter of the frame but slightly down from the "equator." so the measure between those points would be the spine width...... what is it?
ps- using the 16mm width will give the wrong angle as an answer for the effective cutting angle. to solve for cutting angle you NEED to know the width between those points where the frame contacts the hone......

I actually thought about the point where the stone contacts the radius of the frame would be slightly lower than the "equator" but truthfully wasn't sure exactly how to quantify that. Thus I used the dimensions provided with a little figuring to come up with the half angle. I just forgot to multiply by 2.

We're talking -1°or so

I forgot to multiply by two for the included or total angle

Yeah seems like you guys got it right. tan (alpha) = (1/2)SpineWidth/Blade Width gives you the half angle. So tan alpha = 8/30, or alpha = 14.92 or about 30 deg total.
That's pretty fat cause that piece you hone on is huge! Haha. 30 mm is about a 19/16 blade, but with a 5/8 (16 mm) width spine!! That's a width to spine ratio of about 2:1, giving you about double that of a straight razor angle.
Sharptonn is the frameback piece really that big? Hard to tell from the picture. Course I know nothing about microtomes, maybe that's what they should be.

Haha it is funny isn't it??
I think where you're a little off is that you leveled the face of the blade, but the "zero" is actually at the centerline and not the face. So when you measure the angle without the frame back you should get something other than zero. Given that starting point your going to get something more than what you're showing now, prob closer to the predicted 15 deg.
You can get the centerline level by measuring the angle of both faces until they are equal in magnitude.
Or you can trace it on paper and measure with a protractor

I drew a line from the center of the spine to the edge splitting the blade in half. I then drew a line splitting the frame in half perpendicular to the edge. From the intersection you have to sides of a right triangle. Your side opposite and your side adjacent. 8mm and 29 mm respectively. Using the pathagorean theorem we find the hypotenuse to be 30.08 mm which gives us the the distance from the point of contact on the hone to the edge.

Now to find the half angle I used the equation

Sin(x)=opposite/hypotenuse
Where x= halft the cutting angle.

So simplified this is x=sin^-1(8/3.08)
X=15.42°

Being that's the half angle. We multiply by 2 giving us a whopping 30.84°