Shaper speeds?

[pghmyn]

I can't explain why, but I just can't agree with that statement.  Maybe it's just too broad and/or vague.  But I read that as you are saying the only thing that matters is the power of the machine.  It's saying the sharpness of the blade or the cutter doesn't matter...that the geometry of the cutting edge doesn't matter...  That's just not true.  There are more factors than just the power.  

 

 

 

 

That doesn't explain the bike situation.  I'm saying the power required to move the bike is constant (just as the power required to remove a given amount of material) and that the power output of my legs is constant (just like the power output of the machine's motor)...  However, in a low gear I pedal easily but quickly (faster edge velocity of larger diameter cutter)...and in a high gear I struggle to pedal slowly (bogging down with a smaller cutter). 

I think I understand where he is coming from with the bike example, and I think that is from the total energy equation.

 

∆E = ∆K + ∆U = 0

 

Change in energy = change in kinetic energy + change in potential energy = 0

 

K = (1/2) * mass * Velocity^2

U = -mass * gravity * change in height

 

So, simplifying it:

(1/2) * velocity^2 = gravity * change in height

 

Your mass is constant, it drops out.  So, if you are taking away from your KINETIC ENERGY (reducing velocity), you are increasing your POTENTIAL ENERGY (increasing height).

 

Again, been a little while since physics problems were going on my paper, but the idea is still there.

[dwacker]

You can tell by the sound of the cutter wether they are sharp or not dull carbides sound like two wing corrugated. This is an experience  you won't find it on the internet. Metal machining is a mac and cheese to apples comparison. Shaper cutters and their appropriate feed rates are measured and calculated exactly the same as saw blades. It doesn't matter if its a big ass moulder head, a small roundover cutter, your tablesaw, bandsaw, moulder, planer or even jointer . The only difference is the chip load and spindle speeds. 

A great example of a very low hp inexpensive machine is a WH 1.5hp moulder. IIRC they will cut 6" wide moulding. This tiny machine will will cut big deep mould in a single pass all day long with 1.5hp and two cutters. It does this by having the ability to control feed rate and having the right cutter diameter so as to keep the cutter cool enough to not burn the wood and smoke the cutter it has nothing to do with the machines HP. HP allows you to increase feed rates without stalling. Bigger larger diameter cutters cool themselves more effectively so as long as feed rate remains within the machines capabilities the larger cutters will cut better than the smaller ones. 

Again this has nothing to do with HP. If your looking for faster feed rates the you obviously need to increase HP to prevent stalling. If your just using your little machine in a hobby shop you really don't care about feed rate. 

 

HP gives you production speed nothing more. The correlation between HP and machine build is what holds you back with a smaller machine. Typically the lower HP machines will have smaller spindles, and smaller lighter duty spindle cartridges meaning they can't take the bigger heads. If the machines were built the same as the 5hp machines for example you could do nearly the same work just at a much slower pace.

[Dan S]

I can't explain why, but I just can't agree with that statement.  Maybe it's just too broad and/or vague.  But I read that as you are saying the only thing that matters is the power of the machine.  It's saying the sharpness of the blade or the cutter doesn't matter...that the geometry of the cutting edge doesn't matter...  That's just not true.  There are more factors than just the power.

I would go with lost through the post thread, This is a quote form my origonal post.

This page has a simplified example of the equations used to calculate the power required to remove a given volume of material, all you need to know is the "unit power" of the material. I say simplified, as it doesn't take the type of cutter (hss or carbide read sharpness), or tooling geometry into question, but regardless they are both small factors compared to the power of the machine.

What I mean is HSS cutters require less power to make a cut because they are sharper. helical cutter require less power because they are a sheering cut. rake has an effect as well. They have an effect, but like I said they are small factors. So for example, with a really sharp cutter with the proper rake and shearing angle you might get a 10 or 15% increase in performance but your not going to turn a 1.5hp shaper into a 3hp one.

Radius and rprm doesn't effect the material, because they essentially drop out of the power formulas because of the physics, unless we assume the short term effects of inertia.

[Dan S]

I think I understand where he is coming from with the bike example, and I think that is from the total energy equation.

 

∆E = ∆K + ∆U = 0

 

Change in energy = change in kinetic energy + change in potential energy = 0

 

K = (1/2) * mass * Velocity^2

U = -mass * gravity * change in height

 

So, simplifying it:

(1/2) * velocity^2 = gravity * change in height

 

Your mass is constant, it drops out.  So, if you are taking away from your KINETIC ENERGY (reducing velocity), you are increasing your POTENTIAL ENERGY (increasing height).

 

Again, been a little while since physics problems were going on my paper, but the idea is still there.

pretty close

on flat ground the only thing limiting your speed is aerodynamic drag, and the friction loss of your power train. The aro dynamic drag is the big one in this case. get rid of the aero dray or even minimize it and you get insanity like this.

going up hill you have to deal with the force of gravity, and the aerodynamic drag. Thus you slow down, and some of the power you put towards nullifying the aerodynamic drag, goes towards overcoming the force of gravity.

Some genetics come into play hear as well, such as your ratio of fast twitch muscle fibers to slow twitch, Vo2 max etc.

[Vyrolan]

Dan is totally correct with one caveat. There is a place where the power required to make the cut does not overbalance the motor. This may be what inertial energy gains you with a larger diameter cutting wood. The additional cutter speed may overcome the force required to make the wood sheer.

I think you are right and say it far more eloquently.

Sam and Dan, thanks for the further information. I think the bicycle analogy sticks in my brain because I feel like the bigger cutter is the same trade. The larger cutter with its higher velocity at the cutting edge can do the same work more even just like I can pedal more even and smooth in a lower gear.

I also see what you're saying about some of the blade characteristics being a smaller factor. I think those things are pointed out when you're right at the limit of capability because that's when those small factors are most noticeable in your ability to do something or not.

Great discussion! Sorry for my contribution to the hijacking of this thread. =p

[dwacker]

Lets put some of this to use. How much HP would be required?

 

1.25" spindle

4" diameter two knife dual hook cutter head using 12 degree hook angle.

Cutter height 4"

HSS corrugated 4 flute cutter full height

.5" in total depth of cut

.030 finish chip load

10,000 rpm spindle speed

100 fpm feed rate 46 fpm and 20 fpm calculate each

Material 1x4 poplar

[wtnhighlander]

It seems like there is a lot of confusion regarding the calculation of (horse)power requirements to accomplish a given operation. Remember that horspower is (force * distance) / time. What I see is that no one has shiwn how much force must be applied at a given distance to remove a specific amount of material in a given time.

Don's question should provide the necessary data to determine that, but I suspect that it will involve certain values that the industry professionals have determined through observation, and which may not be readily available to us hobbyists.

All physics calculations aside, I'm inclined to trust the statements of the man with the experience.

Oh, and I learned something as I checked my facts. There are imperial hp, and metric hp, and they are different: http://en.m.wikipedia.org/wiki/Horsepower

[Dan S]

Lets put some of this to use. How much HP would be required?

 

1.25" spindle

4" diameter two knife dual hook cutter head using 12 degree hook angle.

Cutter height 4"

HSS corrugated 4 flute cutter full height

.5" in total depth of cut

.030 finish chip load

10,000 rpm spindle speed

100 fpm feed rate 46 fpm and 20 fpm calculate each

Material 1x4 poplar

Am I to read this as a cut 4" high and 0.5" wide? Thus a cross sectional area of 2 in square.

[Dan S]

Don's question should provide the necessary data to determine that, but I suspect that it will involve certain values that the industry professionals have determined through observation, and which may not be readily available to us hobbyists.

All physics calculations aside, I'm inclined to trust the statements of the man with the experience.

I downloaded a free copy of this and it doesn't have wood broken down by species, but instead classifies it as hardwood softwood, plywood and MDF. I would just need to clarify the cut parameters to see if comes close to making sense, as it doesn't have a setting for shapers

http://www.cnccookbook.com/CCGWizard.html

[Vyrolan]

Am I to read this as a cut 4" high and 0.5" wide? Thus a cross sectional area of 2 in square.

That's probably on the low side...it's cutting along an arc... I'm on my phone so can't calculate half the length of an arc created by a line drawn across a 4" circle 0.5" from the tangent point. Some guesstimating math says its like 1.5" though...so it's scraping 6 square inches...I dunno about the depth though to know how much is cut away each rotation of the cutter.

[Dan S]

That's probably on the low side...it's cutting along an arc... I'm on my phone so can't calculate half the length of an arc created by a line drawn across a 4" circle 0.5" from the tangent point. Some guesstimating math says its like 1.5" though...so it's scraping 6 square inches...I dunno about the depth though to know how much is cut away each rotation of the cutter.

that's why I was asking, because the material is supposed to be 1x4 poplar and I was confused by the nomenclature, as a cut that big would be half the board.

[customwood montrose]

Lets put some of this to use. How much HP would be required?

 

1.25" spindle

4" diameter two knife dual hook cutter head using 12 degree hook angle.

Cutter height 4"

HSS corrugated 4 flute cutter full height

.5" in total depth of cut

.030 finish chip load

10,000 rpm spindle speed

100 fpm feed rate 46 fpm and 20 fpm calculate each

Material 1x4 poplar

I would estimate by experience that cut quality is one thing that is left off this formula. Guessing that if this is a backout, a shaper of 10 HP will be needed for the 100 FPM cut and even for the 46 FPM. A 5HP could do the 20 FPM. Getting back to the 1 1/2 HP shaper, with the standard 3/4 spindle, you should be able to do small raised panels, stick and cope and such. Your feed rates are going to depend on how sharp your cutters are like it has already been mentioned. One thing not mentioned, to fast of cut rate, chatter and chipping; to slow, burning or scorching on your wood and tool life shortened dramatically.

What kind of shapes or cuts are you planning on doing? Do you already have the router bits for those cuts? Shaper cutters are meant to last much longer, the carbide tips are meant to be sharpened many times before the tool is no good. So for most people you will never wear it out if you take care of them. There are many different types of tooling for shapers that can range in price from cheap to you should have won the Powerball!

[tim0625]

All I'm gonna say is Don is back!  Good to hear from you!

[dwacker]

Am I to read this as a cut 4" high and 0.5" wide? Thus a cross sectional area of 2 in square.

 

Sure lets go with that.

 

 

 

 

 

I would estimate by experience that cut quality is one thing that is left off this formula.

 

 

No it was not. Its right there in black and white.

 

 

 

Steve you don't get to answer.

[Dan S]

Sure lets go with that.

obtuseness isn't helpful.

[dwacker]

obtuseness isn't helpful.

 

Not being obtuse. You seem to believe a parameter is missing. Im not going to do the exact math so we will just go with your first guess. Its not really going to make much difference if its a little off one way or another. Anyone that actually understands the machines and cutters already has the answer.

[Dan S]

Not being obtuse. You seem to believe a parameter is missing. Im not going to do the exact math so we will just go with your first guess. Its not really going to make much difference if its a little off one way or another.

I find that hard to believe, as you never specified what kind of profile you are cutting. Is it a cove, Ogee, something completely custom etc. Just saying is a cutter head like is not enough information.

http://www.oellasawandtool.com/products/Titan-100mm-shaper-molder-corrugated-head-dual-12{47}20-hook-4-knife-1-1{47}4.html

[dwacker]

I find that hard to believe, as you never specified what kind of profile you are cutting. Is it a cove, Ogee, something completely custom etc. Just saying is a cutter head like is not enough information.

http://www.oellasawandtool.com/products/Titan-100mm-shaper-molder-corrugated-head-dual-12{47}20-hook-4-knife-1-1{47}4.html

 

4 flutes 1/2" depth just like it says.

[Dan S]

4 flutes 1/2" depth just like it says.

the DOC is only one dimension, the other one is equally important.

for example if we assume thew simplest case, 2 rabbet cuts.

A. 1/2" doc that's 1/2" wide = 0.25 square inches of material removed.

B. 1/2" doc that's 1" wide = 0.5 square inches of material removed.

Cut B will require twice as much power as cut A, because it removes twice as much material.

[dwacker]

 

 

Cut B will require twice as much power as cut A, because it removes twice as much material.

 

No but ok. Go 1/2x1/2 still not going to change the answer. Like I said anyone that understands these machines and cutters already has answer.

[Dan S]

No but ok. Go 1/2x1/2 still not going to change the answer. Like I said anyone that understands these machines and cutters already has answer.

it's mind boggling, that you are incapable of understanding junior high level math and science concepts.

[dwacker]

Rather than responding to the childish insult I gave this a day before getting back. First you have to be able to see the forrest through the trees rather than dwell on an single issue. All of the cuts in question are set up to fail anyone that actually understands cutters can see this without doing any rediculas math calculations. HP is very last thing to consider or not at all. Running the cutter at 100 FPM  was the first clue It doesn't matter what shaper you have or how much HP you have this is not going to happen the wood is going to explode. 46 FPM nope not going to happen either regardless of HP. 26 FPM again to fast and spindle is going to burn the crap out out of the poplar. The tool hook angle is wrong for poplar regardless of HP. The knife in itself is ridiculous to run in a single pass, in the real world that moulding is run with four different knives five if you include a backout. With a single pass its run with the equivalent of 5 shapers all being 10hp and if its a single knife machine its not going to feed anywhere close to 100 fpm. Just a word don't confuse single knife machine with having just one knife. Your planer, jointer are good examples of single knife machines if you have straight knives.

Anyone will always be ahead of the game if you can get past the hyperbole and actually learn the why and how. This applies to every machine in your shop.

[Dan S]

Rather than responding to the childish insult I gave this a day before getting back. First you have to be able to see the forrest through the trees rather than dwell on an single issue. All of the cuts in question are set up to fail anyone that actually understands cutters can see this without doing any rediculas math calculations. HP is very last thing to consider or not at all. Running the cutter at 100 FPM  was the first clue It doesn't matter what shaper you have or how much HP you have this is not going to happen the wood is going to explode. 46 FPM nope not going to happen either regardless of HP. 26 FPM again to fast and spindle is going to burn the crap out out of the poplar. The tool hook angle is wrong for poplar regardless of HP. The knife in itself is ridiculous to run in a single pass, in the real world that moulding is run with four different knives five if you include a backout. With a single pass its run with the equivalent of 5 shapers all being 10hp and if its a single knife machine its not going to feed anywhere close to 100 fpm. Just a word don't confuse single knife machine with having just one knife. Your planer, jointer are good examples of single knife machines if you have straight knives.

Anyone will always be ahead of the game if you can get past the hyperbole and actually learn the why and how. This applies to every machine in your shop.

As you said all the cuts are fake (at least you owned up to it). For example the chipload is bs, you can't have the same chipload at 3 different feed rates with the spindle running at the same rpm. Regardless, The power requirement formulas are not designed to tell you if a cut is possible, but how much power a given cut would take. math and science are not a replacement for common sense, but rather a tool that will let you push a given machine to the limit of what it and it's tooling can safely handle.

[Larry Moore]

I have been trying to stay out of this discussion because I have trouble getting my message into words.

But what has not been brought up is the relation of available torque to hp. A router has a hp and torque rating based on spindle rpm. As you increase the cutter diameter it increases the moment arm and starts to stall the motor, as the rpm falls off so does hp and torque which impacts the ability to produce work.

The 1.5 hp shaper is rated at the motor not the spindle. It has a lot more torque available through the pulleys and or gearing than a router. That is why you can run larger dia cutters.

The bike issue is the same your trying to produce a certain amount of work and when the torque required to go up hill increases you have to change the gear ratio to produce it

[dwacker]

I have been trying to stay out of this discussion because I have trouble getting my message into words.

But what has not been brought up is the relation of available torque to hp. A router has a hp and torque rating based on spindle rpm. As you increase the cutter diameter it increases the moment arm and starts to stall the motor, as the rpm falls off so does hp and torque which impacts the ability to produce work.

The 1.5 hp shaper is rated at the motor not the spindle. It has a lot more torque available through the pulleys and or gearing than a router. That is why you can run larger dia cutters.

The bike issue is the same your trying to produce a certain amount of work and when the torque required to go up hill increases you have to change the gear ratio to produce it

 

Feel free to jump in any time.