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Calculating Cutting Power

MillGuy88

Aluminum
Joined
Mar 13, 2024
Location
Saskatoon
I work in a high-mix, low-volume job shop. When it slows down or if I'm waiting for material, I sometimes like to build my own calculators using spread sheets. I've made one for the power requirements for cutting different materials.

I made my calculator based on Sandvik's parameters which uses a specific cutting force (k) in term of pressure (lbs/in^2 or N/mm^2)
I have a couple of hard copy and PDF versions of Sandvik's 2010 Technical Guide, as well as their 2015 Rotating tools catalog with charts for their specific cutting forces. I noticed I cant find any specific cutting force charts or lists on their website or digital catalogs any more.

Kennametal and the Machinery's Handbook calculate power requirement based on "k", the volume of material removed per power unit per minute (in^3/hp/min. or cm^3/kw/min.)

I'm considering rebuilding my calculator based on Kennametal model but can't find a list or chart as comprehensive as the Sandvik specific cutting forces charts that I have.

I'm curious as to how others are calculating power requirements and what resources they are using for their "k" variables.

My other question is what do people figure their spindle efficiency or power loss through their spindle is?
 
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I use "k" and have approximate values memorized for aluminum, soft steel and cast iron. Since there's considerable "real world" stuff going on between spindle motor and actual cutting, all I care about is 1 significant digit. From that and the cutter features (diameter and tooth count, primarily), I get starting points for feed and cross-section area of cut. This is a simple pencil-and-paper calculation. If not under pressure, I sometimes do it in my head. Based on how things actually behave on the machine (there's that "real world" factor going on), I modify from there.
Actually, there's another factor of time involved for "k": cubic inches per minute per horsepower (or your preferred units).
 
The Machinery's Handbook says it's one cubic inch per minute, per horsepower for plain carbon steel with a Brinell hardness of 280-300. Seems a little hard for a plain carbon steel. (2.73 cm^3/kw/min.)

Sandvik says high carbon steel with a Brinell hardness of 300 has a specific cutting force of 290,000 lbs/in^2. (2000 N/mm^2)
 
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When all is said and done, the force required in a cut is related to the shear strength of the material. If you are having trouble finding data for a particular material, start by using a ratio of the shear strength to something you do have.
 
IME there's no point in going to the nth degree, I have too many bars of almost any material!! change their cutting characteristics part way through
 
The reason I'm asking is I like to include hyperlinks to reference material or pages with the proper charts in the calculators I make, so the data I want is just one click away.

I've made a sheet for specific cutting forces I can link to, but I prefer to link to web pages from reputable sources. I can't link to a page in my Machinery's Handbook. I can link to the PDF's I have, but not a specific page in the PDF so I still have to search for it in the PDF.

Just curious about where others get their info from.
 
Sandvik says high carbon steel with a Brinell hardness of 300 has a specific cutting force of 290,000 lbs/in^2. (2000 N/mm^2)
Area makes weird math. According to this a 1 inch wide file would take 290000 pounds of Umph to cut along a 1 inch thick pierce of steel?
 
Area makes weird math. According to this a 1 inch wide file would take 290000 pounds of Umph to cut along a 1 inch thick pierce of steel?
This is from the Sandvik website so it's in metric

"It can be explained as the force, Fc, in the cutting direction (see picture), needed to cut a chip area of 1 mm that has a thickness of 1 mm. The kc1 value is different for the six material groups, and also varies within each group.

The kc1 value is valid for a neutral insert with a rake angle, γ0, = 0°; other values must be considered to compensate for this. For example, if the rake angle is more positive than 0 degrees, the actual kc value will decrease."

 








 
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