# Thread: Less feed requires more power??

1. ## Less feed requires more power??

I was trying to determine how much HP I would need for a given cut. And wondered something. Why is it according to Machinery's Handbook the higher your feed/revolution. The less power is needed?? I saw I needed 4HP for the cuts I was looking at. But figured I could use a smaller feed and reduce the needed power. Why is it, it holds opposite??

I was using the formula

Pc = KpCQW

2. Pc = KpCQW

Kp = .90 for rolled aluminum
C = 1 (For .3mm/rev)**
Q = 2.91096 Q = V/60ƒd
V = 286.512
ƒ = .3048
d = 2mm
W = 1

Pc = 2.61986 Kw or about 3HP

I figured that if I just didn't take as deep of a cut per revolution. I would be able to reduce the HP needed. Right? Or at least I thought. The problem is if I take .15mm/rev C now = 1.2. If you redo the equation. you're now needing 3.14348 Kw. Why do I need more power to take a smaller cut??

3. Originally Posted by Noodles87
.... Why is it according to Machinery's Handbook the higher your feed/revolution. The less power is needed?? ....
I do not have Machinery's Handbook handy so I cannot find out what material they we dealing with. However, I can tell you I know from experience both turning and milling different steel alloys that you can observe the power requirement falling sometimes when the feed is increased.

When machining steel if the conditions are such that the chip peeling off the work remains fairly cool, and by this I mean below a temperature at which the metal starts to discolor, then the steel retains practically 100% of its normal cold strength. However, if the conditions cause very significant heating of the metal coming off so that the chips are oxidized a brown or blue color, or even come off dull red, then the steel has lost a very large portion of its cold strength. I do not have a source handy for how much strength is lost but I think most simple carbon steels lose more than 60% of their cold strength at a temperature a little bit below dull red. You know that when bending steel by heating it you can take it up in temperature without much change in strength then at a certain point it bends very easily. In the machine the force needed to bend the chip and peel it off decreases in a similar manner so less power is needed.

4. Originally Posted by Noodles87
I was trying to determine how much HP I would need for a given cut. And wondered something. Why is it according to Machinery's Handbook the higher your feed/revolution. The less power is needed?? I saw I needed 4HP for the cuts I was looking at. But figured I could use a smaller feed and reduce the needed power. Why is it, it holds opposite??

I was using the formula

Pc = KpCQW
The power per cubic foot does indeed go down, and there's a bunch of places where this is documented. It's called the specific power requirement, i.e. power per volume. 1.5 hp per in^3 in steel, for example, is a specific power rating.

Most of the force in cutting is used to shear the metal (see shear plane/tool angle to cutting force studies), not to bend it. The less area you spend in shear while removing the same volume of material, the less power per cubic inch is required. Basic strength of materials. The force required for bending does go up, but nowhere near enough to equal the power gained by the shear force decreasing.

• Alright. Thanks a lot. I guess that makes sense.

If I were to use the "Optimum" speed and feed for turning aluminum. According to their feed and speed chart for carbide tooling. A speed of 2820 f/m and 36(0.001in)/revolution. That roughly translates to 10k rpms turning 25mm diameter. If I want to turn at 3500 rpm. Do I simply cross multiply and divde to get me my feed? Or is there another way in calculating it?

Example

X 3500
___ = _____
36 10000

X would equal my feed for 3500 r/m. Or is it not linear?

• Cutting speed in feet per minute.
Code:
```RPM = (cs * pi )/(12 * d)
or
RPM = cs * 3.82/d

2820 * 3.82 / .25 = 43089.6 RPM```
As you can see, 2820 is a rather ambitious speed to try and get a 1/4" endmill to run.

Code:
```IPM = RPM * Edges * Load

43089.6 * 3 * .001 = 129.27 IPM```
However, most machines can feed at 130 IPM.

It's rare for most machine shops to be able to spin an endmill over 10k. You don't need to be anywhere near the optimal value to get a good result.

• Sorry, I'm talking about turning.

• Originally Posted by Noodles87
Sorry, I'm talking about turning.
The only thing that changes in turning is the feed rate, because most people don't use balanced/triple/quad toolholders. Flutes goes to 1.

Also, I noticed you gave the diameter in mm and for some reason I read it as .25.

2820 * 3.82 / .984 = 10498

So yeah, you were right.

10498 * 1 * .001 = 10 ipm

Lathes just take the .001" though, you don't need IPM unless you want to program that way. Personally, .001" is kind of light. Depending on the tool you are using, you might not want to drop below .005" per rev. Sharp HSS, carbide, or a ground insert will do .001". A coated insert will not.

• Really? All turning is done with .001in per revolution? According to my handbook the "optimum" turning is "36" which I assumed was 36 times 0.001. There's also an "average" numbers given for turning 6160 aluminum. Which calls for a higher RPM with a slower feed. Which doesnt make sense to me either. I figured this would be a linear thing with feed/speed charts.

• In traditional machining (non-HSM), roughing/stock removal is best accomplished with a lower operating speed to control the temerature while using a high linear feed to remove as much material as possible. For finishing, you can allow the tool to get a little hotter since you won't be there long, which produces a better surface finish. The low feed does the same; the lower the feed, the smaller the ridges in your part are, the better the finish is. I don't know what Machinery's handbook is smoking by saying 36, because my copy is at work and I can't look it up right now. The ideal feed depends on the geometry of the tool and what you're trying to accomplish, not necessarily the work material.

You misunderstood what I meant by saying "the lathe takes .001 directly." What I mean is that lathes are traditionally programmed in feed per revolution, not feed per minute as a milling machine is. The lathe takes .001", .005", .010" etc values directly when in the appropriate mode, and to program in IPM (10, 20, etc) is unusual.

For some turning tasks, I've fed as much as .070" per revolution into the material for roughing a lot of stock down. For finishing, somewhere between .002" and .005" is satisfactory depending on tool geometry and nose radius.

• I thought I undrestood you right. The numbers you were giving, were per revolution. If it was .001 feed rate with an RPM of 1000. Your Z axis(or whichever) would be moving an 1 IPM respectively.

I'm just trying to figure out how much machine power I need to do the roughing cuts. Finish isn't a worry(for roughing) as there will be finishing passes. I wanted a max cutting speed of 3500 so I would be able to run up near 5000 on the smaller diamters. But maybe I'm looking into that too much?

I wanted to first know the machining power needed. To be able to find the forces acting on things and build my machine to handle those forces. At 3500 rpm I would be very happy to be able to cut at .2mm/revolution with a 2mm depth. Sure...I could settle for a depth of 1mm I guess. I could see for a feed of .1mm/revolution too. But if I dont have to, why should I, you know?

• How big (horsepower/swing) is your lathe, what are you trying to cut?

Aluminum is about .33 hp/in^3, steel is around 1.5 hp/in^3.

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