Scaling up Kossel Mini

I plan to build a Kossel Mini as my first printer, but I don't like the limited build space. I am considering scaling it up, as I can get enough OpenBeam for a massive Kossel for only $12 more than the precut kit (I can cut it to size). I have a few questions, however.

Firstly, how is the build plate held in on a standard Kossel? It looks like it might be difficult to put into a scaled-up model.

Second, is it possible to use GT2 belts that aren't closed loops -- bind the ends somehow? If not, where can I get longer GT2 loops? Could I use fishing line instead?

Third, how rigid is the Kossel Mini? Could a significant scale-up cause too much wobble?

And fourth, is there anything else I missed? Any parts that would need to be changed for a scale-up?

I'm planning to do something similar but I'm going to test it out on the smaller size first.

Do a search for Kossel on Thingiverse and you can find some carriers that do not require a looped belt. The place I'm getting my printed parts from (Maker Geek) agreed to make me a few custom items from thingiverse as part of my order for an additional fee. In particular this should help and it's what I'm trying. I figure that by removing the linear rails it will be much easier to scale vertically and I just liked the clean design of the open loop belt part.

Open Loop Belt mod - [www.thingiverse.com]

Carrier mod to not use linear rails - [www.thingiverse.com]

Edited 1 time(s). Last edit at 01/31/2014 11:22AM by WZ9V.

Quote
epicepee
use GT2 belts

Does anybody have information about tensile modulus of GT2 and T2.5?
Based on this post [groups.google.com], you may want to prefer T2.5 belts to GT2 betls for big delta bots.

I believe that the GT2 is the superior belt for precision registry when all factors are taken into consideration.
The GT2 uses a fiberglass cord which is very close in performance to steel.



For the sake of completeness, the three additional belt profiles shown in
Figure 19j, 19k, and19l are used in Europe and are sometimes found on machinery imported from Europe and Japan.
They are not produced in the U.S.A. and are not covered by RMA standards.
The belts are made of polyurethane, and steel is usually used as the tension member.



In Figure 2, the photoelastic pattern shows the stress distribution within teeth of different geometry. There is a definite stress concentration near the root of the trapezoidal belt tooth, with very low strains elsewhere. For the curvilinear tooth, there is a uniform, nearly constant, strain distribution across the belt. The load is largest in the direction of the tension member to which it is transferred. Because of their superior load carrying capabilities, the curvilinear belts are marketed under the name of Gates' HTD drives. This is an abbreviation of High Torque Drives. As a result of continuous research, a newer version of the curvilinear technology was developed by Gates, which was designated as Gates' PowerGrip GT belt drives.

The trapezoidal shape timing belt was superseded by a curvilinear tooth profile which exhibitedsome desirable and superior qualities.
Advantages of this type of drive are as follows:
•Proportionally deeper tooth; hence tooth jumping or loss of relative position is less probable.
•Lighter construction, with correspondingly smaller centrifugal loss.
•Smaller unit pressure on the tooth since area of contact is larger.
•Greater shear strength due to larger tooth cross section.
•Lower cost since a narrower belt will handle larger load.
•Energy efficient, particularly if replacing a "V" belt drive which incurs energy losses due toslippage.
•Installation tension is small, therefore, light bearing loads.

The PowerGrip GT System, featuring a modified curvilinear belt tooth profile, provides timing and indexing accuracy superior to the conventional PowerGrip Trapezoidal Belt System. Plus, PowerGrip GT Belts have a higher capacity and longer belt life than trapezoidal belts.

It's difficult to make a true quantitative comparison between the backlash of a trapezoidal tooth drive and PowerGrip GT drive due to the difference in "pulley to belt tooth" fit.



Steel
Tensile Strength 360,000 lbs/in^2
Elongation at break 2.5%
Modulus (approx.)15,000,000 lbs/in^2

Fiberglass
Tensile Strength 350,000 lbs/in^2
Elongation at break 2.5 – 3.5%
Modulus 10,000,000 lbs/in^2

The most important advantages are:
1.High strength.
2.Low elongation or stretch.
3.Excellent dimensional stability.
4.Excellent chemical resistance.
5.Absence of creep, 100% elongation recovery.

Disadvantages:
1.High modulus (difficult to bend).
2.Brittleness of glass. Improper handling or installation can cause permanent damage.
3.Poor shock resistance. No shock absorbing quality when used in timing belts

Polyester
Tensile Strength 160,000 lbs/in^2
Elongation at break 14.0%
Modulus (approx.)2,000,000 lbs/in^2

One of the main advantages of polyester cord over higher tensile cords is the lower modulus of polyester, enabling the belt to rotate smoothly over small diameter pulleys. Also, the elastic properties of the material enable it to absorb shock and dampen vibration. In more and more equipment, stepping motors are being used. Polyester belts have proven far superior to fiberglass or Kevlar reinforced belts in these applications. High-speed applications with small pulleys are best served by polyester belts under low load.

Kevlar
Tensile Strength 400,000 lbs/in^2
Elongation at break 2.5%
Modulus 18,000,000 lbs/in^2

High tensile strength and low elongation make this material very suitable for timing belt applications.
Kevlar has excellent shock resistance and high load capacity.

T-5 The main stress line in a trapezoidal tooth timing belt is at the base of the teeth. During operation, this stress greatly reduces belt life. The PowerGrip GT system overcomes this condition with its complete tooth flank contact which eliminates the tooth stress line area. This greatly increases belt life and prevents tooth distortion caused by drive torque. In addition, the conventional timing belt has a chordal effect as it wraps small pulleys. This is significantly reduced in the PowerGrip GT system because there is full tooth support along the pulley.

Full support improves meshing, reduces vibration and minimizes tooth deformation. On drives using a low installation tension, small pulleys, and light loads, the backlash of thePowerGrip GT system will be slightly better than the trapezoidal timing belt system. However, with increased tension and/or loads and/or pulley sizes, the performance of the PowerGrip GT system becomes significantly better than the trapezoidal timing belt system.

The PowerGrip GT system is an extension of the HTD system with greater load-carrying capacity. HTD was developed for high torque drive applications, but is not acceptable for most precision indexing or registration applications. The HTD design requires substantial belt tooth to pulley groove clearance (backlash) to perform.As smaller diameter pulleys are used, the clearance required to operate properly is increased. HTD drive clearance, using small diameter pulleys, is approximately four times greater than anequivalent GT timing belt drive.

The PowerGrip GT system's deep tooth design increases the contact area which provides improved resistance to ratcheting. The modified curvilinear teeth enter and exit the pulley grooves cleanly, resulting in reduced vibration. This tooth profile design results in parallel contact with the groove and eliminates stress concentrations and tooth deformation under load. The PowerGrip GT design improves registration characteristics and maintains high torque carrying capability. PowerGrip GT belts are currently available in 2 mm, 3 mm and 5 mm pitches.

Specific advantages of the PowerGrip GT system can be summarized as follows:
•Longer belt life
The strong fiberglass tensile cords wrapped in a durable neoprene body provide the flexibility needed for increased service life. The deep tooth profile provides superior load-carryingstrength and greatly reduces ratcheting when used with pulleys provided by a licensed supplier.

•Precision registration
PowerGrip GT belts provide timing and synchronization accuracy that make for flawless registration, with no loss of torque carrying capacity.

•Increased load-carrying capacity
Load capacities far exceed HTD and trapezoidal belt capabilities making PowerGrip GT belts the choice for accurate registration, heavy loads and small pulleys.

•Quieter operation
The PowerGrip GT belt's specially engineered teeth mesh cleanly with pulley grooves to reduce noise and vibration. Clean meshing and reduced belt width result in significant noise reduction when compared to Trapezoidal and HTD belts.

•Precise positioning
PowerGrip GT belts are specifically designed for applications where precision is critical, such as computer printers and plotters, laboratory equipment and machine tools.
[www.scribd.com]

REFERENCE LIBRARY
[www.sdp-si.com]

T5 slightly stiff so it requires a higher tension
[forums.reprap.org]

Belts and Pulleys
[www.reprap.org]

Edited 1 time(s). Last edit at 02/01/2014 02:16AM by A2.

Edits in italic.

Nice info A2. Unfortunately I believe insufficient to decide whether to go with glass reenforced belt or steel reenforced belt.

OK, it is clear that curvilinear tooth profile is better than trepezoidal. I do not have doubts about that.

The problem of using fiber glass instead of steel chords is different. I do not think it is as clear. I did a very rough computation for a T2.5 belt I have here. It's steel crosssection area is about 0.28 mm². A typicall stepper dynamical force change (usable torque about 0.2 N/m and pulley with 14 mm diameter) can be about 2 * 0.2 / (0.014/2) = 57N. Young modulus of steel (based on A2's reference) is about 103.4 GPa. (Other references go up to 200 GPa). Lets approximate how much the belt would stretch per 1 metre of its length:
That means 103.4 = 57 * 1 / (0.28e-6 * ∆l)
From that: ∆l ≅ 0.002 m mm = 2 mm.
If fiber glass would be used instead of steel, we would get 0.003 m mm = 3 mm.

These are pretty good results. It would indicate we should definitely use GT2 belts.
Actualy the results are not good. They indicate the belt stretching can be a significant problem. A few milimeters dynamical error is big. The longer the belts the worse it will be. There is a chance the maximum stretching estimation is too pesimistic though. Maybe I underestimated the cross section of the steel wires in the belt I have at hand. We can also hope the motors are not running near their torque limits. It was also the worst case when we accelerate in one direction at the stepper skip limit and then accelrate the oposite direction again at the stepper skip limit. This cannot happen with the right acceleration/jerk settings in firmware. Intertia of the stepper rotor will limit the amount of force change at the belt too. Anyway I believe this indicates that belt dynamic elongation cannot be dismissed without consideration.

From other point of view I think the results significantly understimate the stretching problem. The reason is that I estimated it for a signle strand of steel/glass. The belt will contain multiple strands which will be twisted to a string. And then there would be multiple strings in a belt. I believe the young modulus for such a twisted string will be worse (smaller) than for one fiber only. A string of multiple twisted strands should be significantly more stretchy than one strand. More over the glass reinforced belt will have more finer fibers because glass is brittle. So it should be worse for it.

To really know, we need young modulus for the belts themselves not for the base material used.

Edited 1 time(s). Last edit at 02/01/2014 03:42PM by hercek.

Quote
hercek
Unfortunately I believe insufficient to decide whether to go with glass reenforced belt or steel reenforced belt.

I also think they also significantly understimate the stretching problem.

If your load is well under the young modulus you won't have "meaningful" stretch.

PowerGrip GT2
There is minimal stretching during the wear-in period.
After the burn in period there is no stretching due to wear.

The effects of belt elongation and tooth deflection do not have any influence on the registration accuracy of this type of system.
[www.cad.sun.ac.za]

Timing Belt Theory
pg13
the total elongation (deformation) of the belt operating under load is equal to the total belt elongation resulting from the belt pre-tension.

For most practical cases the difference between the deformations of the belt in contact with both pulleys during pretension and during operation is negligible.

Tensile tests show that in the tension range timing belts are used, stress is proportional to strain.

Theoretically, the tooth stiffness increases with increasing belt tension over the tooth, which has also been confirmed empirically.
This results in the practical recommendation for linear actuators to operate under high pre-tension in order to achieve higher stiffness, and hence, better positioning accuracy.
[www.gatesmectrol.com]

The fiberglass tensile member provides greater length stability than competitive belts using aramid tensile members.
[www.cad.sun.ac.za]

... in the first few hours of operation of a new belt, the belt takes a small permanent stretch, which has the effect of reducing the preload.
So the preload values for a new belt are higher than for a used belt,

The amount of force to produce a specified stretch is known as the belt modulus.
[www.epi-eng.com]

Young modulus
(10^9 N/m2, GPa)
Glass 50 - 90.
Steel, Structural ASTM-A36: 200.
[www.engineeringtoolbox.com]

Young modulus
Young modulus is defined as the ratio of the stress along an axis over the strain along that axis in the range of stress in which Hooke's law holds.
[en.wikipedia.org]

Hooke's law
It must eventually fail once the forces exceed some limit, since no material can be compressed beyond a certain minimum size, or stretched beyond a maximum size, without some permanent deformation or change of state.
[en.wikipedia.org]

ISO 5294:1989: Synchronous belt drives -- Pulleys
ISO 5296-1:1989:1989: Synchronous belt drives -- Belts -- Part 1: Pitch codes MXL, XL, L, H, XH and XXH -- Metric and inch dimensions

Timing Belts for 3D Printers
Request a copy.
[www.fennerprecision.com]

Edited 2 time(s). Last edit at 02/01/2014 03:58PM by A2.

Quote
A2
If your load is well under the young modulus you don't have "meaningful" stretch.

Well, as many times as it is below the young modulus, that many times the stretch is less meaningfull for unit length of belt. Anyway this is meaningless claim without some concrete numbers/examples in comparison to our delta bots. At least rmat claims dynamic stretch is a problem. Maybe he bought low quality belts or pulleys. Or maybe it is a real problem.

Quote
A2
PowerGrip GT2
There is minimal stretching during the wear-in period.
After the burn in period there is no stretching due to wear.

The effects of belt elongation and tooth deflection do not have any influence on the registration accuracy of this type of system.
[www.cad.sun.ac.za]

That is about static stretch during wear-in. That is not what I'm concerned about. I believe that is not what rmat was talking about. I was not able to find modulus (or the linear stifness) of the belt itself in the linked PDF. That is what I'm interested in. How does it compare between GT2 (fiber glass reinforcement) and T2.5 (steel reinforcement).

Quote
A2
Timing Belt Theory
pg13
the total elongation (deformation) of the belt operating under load is equal to the total belt elongation resulting from the belt pre-tension.

For most practical cases the difference between the deformations of the belt in contact with both pulleys during pretension and during operation is negligible.

Tensile tests show that in the tension range timing belts are used, stress is proportional to strain.

Theoretically, the tooth stiffness increases with increasing belt tension over the tooth, which has also been confirmed empirically.
This results in the practical recommendation for linear actuators to operate under high pre-tension in order to achieve higher stiffness, and hence, better positioning accuracy.
[www.gatesmectrol.com]

Note that on page 16 of the paper they explicity take belt stiffness Kr into account when computing the total drive stiffness. So clearly dynamic belt stretching (I do not mean the initial wear-in stretching) needs to be taken into account for precise positioning.

Quote
A2
The amount of force to produce a specified stretch is known as the belt modulus.
[www.epi-eng.com]

Yes, belt modulus for both GT2 and T2.5 is the number I would like to see. Then I would be able to guess whether rmat might have been right with his claim that GT2 is not good since it is too stretchy.

Maybe somebody who has both GT2 (fiberglass) belt and T2.5 (steel) belt can measure this for us?

Belt tensile properties: [www.bbman.com]

TheTechnicalNoob: thanks for the link.

Hell, that documents must be writen by heretics of physics. Not only the authors use "hogshead" units, but they cannot even get it right, even with them. Ok, so they define modulus as lb/in² but (based on the formula (in note 7) for ussage of their modulus values) the unit is actually only lb. Or at least I hope this is what they intended (one cannot be completely sure when they have contradictions even in such a simple document).

Ok, so for GT2 belt it would be 18000 lb for 1" wide belt. Based on note 4, that is 18000/25.4*6*0.82 ≅ 3486.6 lb (or about 15509 N) for the common 6mm belt repraps often use. That means the elongation for 1m long belt and 57N force change is 1 * 57 / 15509 ≅ 0.0037 m = 3.7 mm.
That is quite a lot but suprisingly near to the value I estimated before for the glass core (which was 3mm). Probably just a luck since I was only guessing what the glass core cross section may be. Anyway, it is good sign that it does not differ by order or more orders of magnitude.

That indicates that young modulus may not be that different for a string of twisted fibers compared to a single fiber. That indicates that the individual filaments are barely twisted around themselves when spun to the reinforcement string of a belt and/or that a good pretension is applied when the belt is produced.

Anyway, 3.7mm per metre is a lot! Even if we decide that printers typically use only 1/10 of the needed stepper torque when changing direction of movement, even with this assumption we get 0.37 mm tower position error which will lead to about the same errors in the effector movement. And only due to the belt elasticity. That also means that printed part errors we see around corners may be also because of belt elasticity (and not only because of the surface tension and contraction of the cooling plastic).

Now we only need the same data for the belts with a steel core. They should be better by at least 1/3.

Edited 1 time(s). Last edit at 02/02/2014 05:39PM by hercek.

@hercek:

I found no contradictions in the Gates chart.
Post the units, and show your work to avoid confusion.

Using a 6 mm wide GT2 belt with a 28.57 newton-force (6.42 pound-force),
there is 0.00023 mm (0.000009 inch) of belt stretch.
For all practical purposes, this is equivalent to zero stretch.

For All Practical Purposes (FAPP)
FAPP is a pragmatic approach towards the problem of incompleteness of every scientific theory and the usage of asymptotical approximations.
[en.wikipedia.org]

Quote
hercek
A typicall stepper dynamical force change (usable torque about 0.2 N/m and pulley with 14 mm diameter) can be about 2 * 0.2 / (0.014/2) = 57N.

0.20 N-m holding torque.
14 mm diameter pulley.

1 meter = 1000 mm
1 millimeter = 0.001 meter

14 mm dia / 2 = 7 mm radius
7 mm / 1000 mm/m = .007 meter
7 mm radius * 0.001 meter = .007 meter

0.2 N/m / .007 m = 28.57 N
28.57 newton = 6.423 pound-force

Young's modulus has units of pressure.
Young's modulus = Tensile modulus or elastic modulus.
[en.wikipedia.org]

Pounds per square inch
[en.wikipedia.org]

Quote
hercek
18000/25.4*6*0.82 ≅ 3486.6 lb (or about 15509 N)

1 inch² = 0.00064516 meter²
1 pound-force = 4.4482 newton

Tensile Modulus = 18000 pound-force/inch²
18000 pound-force = 80068 newton
Tensile Modulus = 80068 newton/0.00064516 meter²
[www.translatorscafe.com]

Belt Elongation = (Belt Span Length x Tensile Load) / Tensile Modulus
Belt Elongation = (1.0 meter * 28.57 N) / 80068 N/0.00064516 meter²
Belt Elongation = (1.0 meter * 28.57 N) / 124105648 N/meter²
Belt Elongation = 0.00000023 meter = 0.00023 mm = 0.000009 inch

Edited 2 time(s). Last edit at 02/03/2014 04:31AM by A2.

Quote
A2
I found no contradictions in the Gates chart.
Post the units, and show your work to avoid confusion.

Claim 1 of the document:
The last column of the table specifies unit as lb/in² for 1" wide belt.
Claim 2 of the document:
Note 7 specifies the formula for the belt elongation computation as BeltElongation = (BeltLength * TensileLoad) / TensileModulus.
Lets specify what should be the units for all the terms of this equation except the TensileModulus:

• for BeltElongation it is in
• for BeltLength it is in
• for TensileLoad it is lb

Lets put the units into the equation and derive the unit for their TensilaModulus:
in = (in*lb)/TensileModulus
TensileModulus = (in*lb)/in
TensileModulus = lb
That means that from the Note 7 we can deduce that their unit for TensileModulus is actually lb.

Claim 1 is different from Calim 2 because lb ≠ lb/in².
That is the contradicition in the document. The problem with contradictions is that anything can be deduced from them. I interpreted it one way and deduced elongation of about 1.85 mm (If I would assume load of only 29N instead of 57N), you interpreted it another way and deduced elongation of 230 nm which is about a wavelength of ultraviolet light. What interpretation sounds more probable?

Quote
A2
0.2 N/m / .007 m = 28.57 N
28.57 newton = 6.423 pound-force

I multiplied 28.57N by 2 and got about 57N becasue the stepper can produce the force of 28.57N (ignoring the stepper rotor inertial forces) in one direction and (with big enough jerk) in the opposite direciton too just a moment later. So once we have force on the belt of +28.57N and the moment later -28.58N. The difference is about 57N. I did it this way to get the worst ever possible situation. If the worst possible situation would lead to neglible elongation then I would know this is not something to be concerned about. But I easily grant you that we should use number 28.57N ... really this does not matter much. My computation ingnores so many things that factor of 2 is almost nothing compared to other possible errors. But I would start to be concerned at about factor of 10 and more.

As for as the rest of your computaion. It is correct. You just selected the other interpretaion of the document with condtradictions.
Except the last step. You have the last step wrong. If you would bother to continue to folow also the units in the very last step you would even notice the contradiction in the document since:

Quote
A2
Belt Elongation = (1.0 meter * 28.57 N) / 124105648 N/meter²
Belt Elongation = 0.00000023 meter = 0.00023 mm = 0.000009 inch

Belt Elongation = (1.0 meter * 28.57 N) / (124105648 N/meter²)
Belt Elongation = 230e-9 * (meter*N) / (N/meter²)
Belt Elongation = 230e-9 * meter / (1/meter²)
Belt Elongation = 230e-9 * meter³
Ooops, belt elongation in cubic meters? Does not sound right to me.

Anyway, in the absence of more data, I'm tempted to think that my interpretation of the contradictory document is the correct one. Especialy because it is about the same as the estimation I did here for my T2.5 belts with steel core. There can be a big error in my estimation because I more or less guessed the steel core filament diameter in my belt (there was no easy way to measure it without cutting off and dismantling a piece of the belt). But I doublt I guessed it wrong by 3 orders of magnitude.

Uff, I do not like imperial units. They are a mess. And I do not have experience with them. And I do not even want the experience. People who use them (like the authors of the document) should at least use them right so that they do not confuse the hell out of us SI users who want to keep it simple.

Hmm, I think the best option how to built an argument that GT2 belts are good enough would be to try to argue that the moment of inertia of stepper motor rotor is so big compared to the inertia of the carriage/platform/hotend/fan that the actuall dynamic forces on the belt will be too small to lead to any significant elongation. My hunch is that neglecting this is the biggest source of error in my computation. I do not have a guess how big it is without putting down some equations though.

Quote
hercek

Quote
A2
I found no contradictions in the Gates chart.
Post the units, and show your work to avoid confusion.

Claim 1 of the document:
The last column of the table specifies unit as lb/in² for 1" wide belt.
Claim 2 of the document:
Note 7 specifies the formula for the belt elongation computation as BeltElongation = (BeltLength * TensileLoad) / TensileModulus.
Lets specify what should be the units for all the terms of this equation except the TensileModulus:

• for BeltElongation it is in
• for BeltLength it is in
• for TensileLoad it is lb

Lets put the units into the equation and derive the unit for their TensilaModulus:
in = (in*lb)/TensileModulus
TensileModulus = (in*lb)/in
TensileModulus = lb
That means that from the Note 7 we can deduce that their unit for TensileModulus is actually lb.

Claim 1 is different from Calim 2 because lb ≠ lb/in².
That is the contradicition in the document. The problem with contradictions is that anything can be deduced from them. I interpreted it one way and deduced elongation of about 1.85 mm (If I would assume load of only 29N instead of 57N), you interpreted it another way and deduced elongation of 230 nm which is about a wavelength of ultraviolet light. What interpretation sounds more probable?

Quote
A2
0.2 N/m / .007 m = 28.57 N
28.57 newton = 6.423 pound-force

I multiplied 28.57N by 2 and got about 57N becasue the stepper can produce the force of 28.57N (ignoring the stepper rotor inertial forces) in one direction and (with big enough jerk) in the opposite direciton too just a moment later. So once we have force on the belt of +28.57N and the moment later -28.58N. The difference is about 57N. I did it this way to get the worst ever possible situation. If the worst possible situation would lead to neglible elongation then I would know this is not something to be concerned about. But I easily grant you that we should use number 28.57N ... really this does not matter much. My computation ingnores so many things that factor of 2 is almost nothing compared to other possible errors. But I would start to be concerned at about factor of 10 and more.

As for as the rest of your computaion. It is correct. You just selected the other interpretaion of the document with condtradictions.
Except the last step. You have the last step wrong. If you would bother to continue to folow also the units in the very last step you would even notice the contradiction in the document since:

Quote
A2
Belt Elongation = (1.0 meter * 28.57 N) / 124105648 N/meter²
Belt Elongation = 0.00000023 meter = 0.00023 mm = 0.000009 inch

Belt Elongation = (1.0 meter * 28.57 N) / (124105648 N/meter²)
Belt Elongation = 230e-9 * (meter*N) / (N/meter²)
Belt Elongation = 230e-9 * meter / (1/meter²)
Belt Elongation = 230e-9 * meter³
Ooops, belt elongation in cubic meters? Does not sound right to me.

Anyway, in the absence of more data, I'm tempted to think that my interpretation of the contradictory document is the correct one. Especialy because it is about the same as the estimation I did here for my T2.5 belts with steel core. There can be a big error in my estimation because I more or less guessed the steel core filament diameter in my belt (there was no easy way to measure it without cutting off and dismantling a piece of the belt). But I doublt I guessed it wrong by 3 orders of magnitude.

Uff, I do not like imperial units. They are a mess. And I do not have experience with them. And I do not even want the experience. People who use them (like the authors of the document) should at least use them right so that they do not confuse the hell out of us SI users who want to keep it simple.

Young's modulus has units of pressure (N/m2).
[en.wikipedia.org]

Add/Subtract/Multiply/Divide Fractions with Exponents, (video):
[www.youtube.com]

Quote
hercek
Lets put the units into the equation and derive the unit for their TensilaModulus:
in = (in*lb)/TensileModulus
TensileModulus = (in*lb)/in
TensileModulus = lb
That means that from the Note 7 we can deduce that their unit for TensileModulus is actually lb.

Belt Elongation = (1.0 meter * 28.57 N) / (124105648 N/meter²)
Belt Elongation = 230e-9 * (meter*N) / (N/meter²)
Belt Elongation = 230e-9 * meter / (1/meter²)
Belt Elongation = 230e-9 * meter³
Ooops, belt elongation in cubic meters? Does not sound right to me.

I fail to understand your reasoning as to why my calculation is incorrect, and how you derived cubic meters.
If my math is in error, hopefully someone can help me out with it.

Quote
hercek

Anyway, in the absence of more data, I'm tempted to think that my interpretation of the contradictory document is the correct one.
... And I do not even want the experience.

Relativism is the concept that points of view have no absolute truth or validity,...
[en.wikipedia.org]

Quote
hercek
I multiplied 28.57N by 2 and got about 57N becasue the stepper can produce the force of 28.57N (ignoring the stepper rotor inertial forces) in one direction and (with big enough jerk) in the opposite direciton too just a moment later.

Jerk?
[en.wikipedia.org])

Oops, I found a notational error, HTH?
0.2 N/m / .007 m = 28.57 N
0.2 N-m / .007 m = 28.57 N

International System of Units
[en.wikipedia.org]

(British) Imperial units
[en.wikipedia.org]

Quote
hercek
I do not have a guess how big it is without putting down some equations though.

I'm looking forward to your equations.

Have you had the time to review the links that I've posted?

zsh> maxima
Maxima 5.32.1 [maxima.sourceforge.net]
using Lisp SBCL 1.1.14
Distributed under the GNU Public License. See the file COPYING.
Dedicated to the memory of William Schelter.
The function bug_report() provides bug reporting information.
(%i1) A: 1.0 * meter * 28.57 * Newton / (124105648 * Newton / meter^2);
(%o1)                     2.302070893663115e-7 meter^3
(%i2) ^D
zsh>

If you still do not believe it then use the function bug_report().

@hercek:

I installed wxMaxima 13.04.2 software.
Your argument is incorrect.



2.3020708936631152*10^-7 meters
0.00000023020708936631152 meters
0.00023020708936631152 mm

Quote
A2
Using a 6 mm wide GT2 belt with a 28.57 newton-force (6.42 pound-force),
there is 0.00023 mm (0.000009 inch) of belt stretch.
For all practical purposes, this is equivalent to zero stretch.

0.00023 mm is exactly the same result that I posted earlier.

Try the following script.

A: 1.0 * meter * 28.57 * Newton ;
B: 124105648 * Newton * meter^2 ;
C: A / B ;

Edited 2 time(s). Last edit at 02/04/2014 01:34AM by A2.

A2, I needed a break from this discussion. But I'll give you one more chance. Maybe you made only a mistake (maybe you are not trolling me).

1. Why should I use a script which is different from mine and from the one specified in the Note 7 of the Gates document. In your script, the definition of B should have been: 124105648 * Newton / meter²; (Because the unit for their "modulus" was lb/in². Because you changed the problem without any explanation why you are doing so, it looks like you are trying to pull out a Straw Man.
2. Even If I would try to use your script (exactly as you have written it), notice that you got the final unit of elongation (meter) in the position of a denominator. That means that you are actually proposing to measure distances in units of 1/meter. Does it not sound bad to you? Distances are typically measured in units of meter, not 1/meter.

Quote
hercek
I interpreted it one way and deduced elongation of about 1.85 mm

Gates claims their GT2 belt has zero stretch, you "deduced" that the stretch is 1.85 mm.
That's a lot of stretch for a 6.423 pound-force.
To put this in perspective, you could easily test your deduction simply by tensioning the belt with your hands.
Place two marks on the belt 1 meter apart, and measure how much they separate.

Quote
A2
0.2 N/m / .007 m = 28.57 N
28.57 newton = 6.423 pound-force

Quote
hercek
A2, I needed a break from this discussion. But I'll give you one more chance. Maybe you made only a mistake (maybe you are not trolling me).
Why should I use a script which is different from mine and from the one specified in the Note 7 of the Gates document. In your script, the definition of B should have been: 124105648 * Newton / meter²; (Because the unit for their "modulus" was lb/in². Because you changed the problem without any explanation why you are doing so, it looks like you are trying to pull out a Straw Man.

I show the math of the Gates formula, and I provided you a link to help you with the modulus conversion.
You can check your hand calculated modulus value against the on-line calculator.

Quote
hercek
Even If I would try to use your script (exactly as you have written it), notice that you got the final unit of elongation (meter) in the position of a denominator. That means that you are actually proposing to measure distances in units of 1/meter. Does it not sound bad to you? Distances are typically measured in units of meter, not 1/meter.

You don't divide your answer by the distance.
The output is the length that the belt will stretch per 1 meter.
That's why the wxMaxima software shows you the result over the span of interest.



Edited 1 time(s). Last edit at 02/06/2014 04:06PM by A2.

Uff, ok, I took apart the rostock here and I measured the T2.5 belt with steel core. I pre-tensioned the 1.48m long belt with force of 190N. Then I needed to increase the force to 280N to make it longer by about 2.5mm (one belt teeth span). That corresponds to elontagion of about 0.02mm per 1 meter of belt and 1 newton of force. The glass core GT2 must be super hight quality to have this constant of about 230nm / 28.57N ≅ 8nm. Sure my measurement is approximate and it is a steel core belt, but the difference is about 3 orders of magnitude!

Three last questions for you. You do not need to answer. I'm putting them here to make you see the problem. I assume you always tie unit to the corresponding constatnt with the hightest priority. Otherwise a lot of parantheses are missing in your picture. Citation from your picture:
Tensile Modulus = 80068 newton/0.00064516 meter²
Belt Elongation = (Belt Span Length x Tensile Load) / Tensile Modulus
I modify the last line of the citation by adding paranteses (I hope you believe it is semantically the same):
Belt Elongation = (Belt Span Length x Tensile Load) / ( Tensile Modulus )
Now I replace string Tensile Modulus with its right hand side from the first line of the citation:
Belt Elongation = (Belt Span Length x Tensile Load) / ( 80068 newton/0.00064516 meter² )
Another citation from your picture:
Belt Elongation = (Belt Span Length x Tensile Load) / 80068 newton/0.00064516 meter²
Do you think the last two bold lines of this post are semantically the same?
Which line corresponds to the script you entered in Maxima?
Which line correspondst to the script I entered in Maxima?

For me to interpret your comments, and provide meaningful commentary,
you will need to be more rigorous with your language, and show your math.

Quote
hercek
Uff, ok, I took apart the rostock here and I measured the T2.5 belt with steel core.
I pre-tensioned the 1.48m long belt with force of 190N.
Then I needed to increase the force to 280N to make it longer by about 2.5mm (one belt teeth span).
That corresponds to elontagion of about 0.02mm per 1 meter of belt and 1 newton of force.

Where is the link to the specification, what size belt, who made it, are you measuring the elongation of 1 meters?

Did you mean:
A 1.0 meter span, of a T2.5 belt, with a 280 N tensile force (62.95 pound-force) elongated by 2.5 mm?

Quote
hercek
That corresponds to elontagion of about 0.02mm per 1 meter of belt and 1 newton of force.

This does not make sense, show your math.

This is the data sheet from Gates [www.bbman.com].
You should compare your empirical results to the T2.5 belt elongation specifications,
you should get the same values, if not you're doing something wrong.

Quote
hercek
Which line corresponds to the script you entered in Maxima?

FYI: this is 5th grade math.

A: 1.0 * meter * 28.57 * Newton ; = (Belt Span Length x Tensile Load) = (1 meter * 28.57 N) = 28.57 N-m.
B: 124105648 * Newton * meter^2 ; = ( Tensile Modulus ) = 124105648 N/meter².
C: A / B ; = 28.57 N-m / 124105648 N/meter² = 2.30207089e^-7 meter = 0.00000023 meter = 0.00023 mm = 0.000009 inch belt elongation.

C: A / B ;
The (m) in the numerator is canceled out by one of the (m) in the denominator.
The (N) in the numerator is also canceled out by the (N) in the denominator leaving you with: 28.57 / 124105648 meter = 2.30207089e^-7 meter.

Tensile Modulus = 18000 pound-force/inch² = 80068 newton/inch²
Proportionally convert 1-inch² into meter², this gives you 80068 newton/0.00064516 meter², as 1 inch² = 0.00064516 meter².

Then to get rid of the decimal area of 0.00064516 meter², you increase the fractional meter² (0.00064516 meter²) to a full 1-meter²,
This then necessitates that you increase the force (80068 N) proportional to the larger area of 1 meter², (i.e. 80068 N to 124105648 N).
Tensile Modulus = 80068 newton/0.00064516 meter² = 124105648 N/meter²

Belt Elongation = (Belt Span Length x Tensile Load) / Tensile Modulus
Belt Elongation = (1.0 meter * 28.57 N) / 80068 N/0.00064516 meter²
Belt Elongation = (1.0 meter * 28.57 N) / 124105648 N/meter²
Belt Elongation = 0.00000023 meter = 0.00023 mm = 0.000009 inch

H.T.H.

Quote
A2
Did you mean:
A 1.0 meter span, of a T2.5 belt, with a 280 N tensile force (62.95 pound-force) elongated by 2.5 mm?

Quote
hercek
That corresponds to elontagion of about 0.02mm per 1 meter of belt and 1 newton of force.

This does not make sense, show your math.

I do not have specification of the belt. The seller did not specify the producer. The Gates document does not mention T2.5 belts so I cannot use that as an approximation. (Regardless we do not agree how to interpret the Gates document.) It was a T2.5 belt. Its width was 6mm. Its length was 1.480m when it was tensioned with force of 190N. Then I increased the force to 280N. That made the belt to stretch to the final length of 1.4825m (1.4825m - 1.480m = 0.0025m = 2.5mm). So how much would this belt stretch if it was 1m long and the applied force was 1N? For that we need to divide the measured elongation of 2.5mm by the force difference and the length of the belt.
2.5 / (280-190) / 1.48 ≅ 0.02
So if it would be 1m long and the force was 1N then the belt would elongate by 0.02mm.

Quote
A2
FYI: this is 5th grade math.

Yes, it is about 5 grade math and you are doing it wrong.
I show you the first error. The rest does not make sense to follow since it is already wrong after the first error:
Here is your expression B:

Quote
A2
B: 124105648 * Newton * meter^2 ; = ( Tensile Modulus ) = 124105648 N/meter².

Since the equivalence relation is transitive we can deduce (I assume N stands for Newton):
124105648 * Newton * meter² = 124105648 * Newton/meter²
Now we can devide by 124105648*Newton:
meter² = 1/meter²
Now we can multiply by meter²:
meter⁴ = 1
Which is a contradiction.
I proofed by contradiction that what you have writen for expression B is not correct.

@ hercek:

I'm aware that there are more involved equations to calculate elongation,
but there is no chance that I'm going to introduce you to higher math, no way!

I did find another formula that is simple to follow, and adds another variable.

You are for some unknown reason having difficulty with unit cancellation.
So I don't hold much hope that you will agree or understand this one either.



BELT Technologies, Inc. equation for Belt Stretch of Steel
[www.belttechnologies.com]

L = (P*L)/(A*E)

Where:
L = Stretch in inches = 0.000006 inch.
P = Tension load in pounds = 6.423 pound.
L = Initial belt length in inches = 39.37 inch.
A = Belt cross - section area in inches = 0.28 millimeter² = 0.000434 inch².
E = Young’s Modulus = 18000 psi.

Quote
hercek
steel crosssection area is about 0.28 mm²

Since we don't know the cross sectional area of the GT2 belts glass fibers, I'll use your value of 0.28 millimeter².



The only way you can get the units to work out using the new input of cross-sectional area is to proportionalize
the area of the glass fiber bundles over 1 square inch before multiplying by the modulus of the belt, i.e. (1* inch * inch /(0.000434 * inch * inch)).
If you don't do that you get meter3, which is a bogus conclusion.

Belt elongation = 0.000006 inch, which is less than my calculation of 0.000009 inch.

The difference is only 0.000003 inch, which is for all practical purposes is still zero belt elongation.
0.000009 inch - 0.000006 inch = 0.000003 inch.

Quote
hercek
Since the equivalence relation is transitive we can deduce (I assume N stands for Newton):
124105648 * Newton * meter² = 124105648 * Newton/meter²
Now we can devide by 124105648*Newton:
meter² = 1/meter²
Now we can multiply by meter²:
meter⁴ = 1
Which is a contradiction.
I proofed by contradiction that what you have writen for expression B is not correct.

You are confusing wxMaxima place holder (*) as multiplication.
If you were given a value of 10 psi, would you try to reduce it?
That's what the modulus is psi, you don't reduce it.

I get the same result every time, I've yet to see you calculate belt stretch, and you are struggling with unit cancellation,
and you continue to make strange transitive deductions that make no sense.

hercek, you continue to claim that the Gates equation is wrong, then why don't you show us the correct equation?

Empirical results are in this case the best that you, or I could expect for.
Go buy a GT2 belt, and make a video of your experiment for all to enjoy.

I eagerly await your correction to the Gates equation.

Edited 1 time(s). Last edit at 02/07/2014 05:34PM by A2.

A2,
Your answer should be in inches, not per inch, so you have the units wrong.

stretch = (P*L)/(A*E) = pounds * inch / (inch^2 * pounds/inch^2) = inch.

Your modulus for glass fibre seems miles out. From here I get 12.3282 x 10^6 PSI.

With those figures I get (6.423 * 39.37) / (0.000434 * 12328200) = 0.0473 inch or 1.2mm, which seems a lot, but not over 1 metre.

Edited 1 time(s). Last edit at 02/08/2014 06:24AM by nophead.

---

[www.hydraraptor.blogspot.com]

Finally some one comes to help!
Tks for pointing that out nophead!

Using wxMaxima I couldn't code it to get the units to work out.
I used your modulus, so you can follow.



The result is in inch^3
Am I interpreting or coding it incorrectly ?

The modulus comes from Gates.
[www.bbman.com]

Edited 1 time(s). Last edit at 02/08/2014 06:43AM by A2.

Maxima is a fairly complete computer algebra system written in Common Lisp with an emphasis on symbolic computation.
It is based on DOE-MACSYMA and licensed under the GPL.
Its abilities include symbolic integration, 3D plotting, and an ODE solver.

[maxima.sourceforge.net]

[sourceforge.net]

The mistake you are making is modulus in PSI is pounds per square inch, not pounds times square inches.

When you multiply PSI by inch squared in the denominator the inch squares cancel leaving just pounds. So the numerator is inch pounds and the denominator pounds giving a result which is just inches.

---

[www.hydraraptor.blogspot.com]

Edits in strike-through and italic.

Quote
A2

The only way you can get the units to work out using the new input of cross-sectional area is to proportionalize
the area of the glass fiber bundles over 1 square inch before multiplying by the modulus of the belt, i.e. (1* inch * inch /(0.000434 * inch * inch)).
If you don't do that you get meter3, which is a bogus conclusion.

You are confused so much here that it is starting to be funny.
Do you know what psi means? It means pounds per squre inch. The word "per" indicates division.
Notice how it is defined in the wiki page here: [en.wikipedia.org]
Notice it is written as lbf/in². Notice there is division there not a multiplication.
How could you write it in maxima as (18000 * pound * inch * inch) instead of the correct (sans the 18000 tern) way which is (18000 * pound / inch^2)?
Instead of writing this correctly into maxima you made up a scifi theory that the term for belt cross section area needs to be somehow proportionalized instead of just used directly with its unit of inch². The expression for area should be (0.000434*inch*inch) and not your magic (1* inch * inch /(0.000434 * inch * inch)). If something in math is not working out (e.g. your final unit in 1/meter³), you do not go to introduce fudge factros willy-nilly to correct for it. You must correct the primary source of error which in your case was writing psi as pound*inch*inch instead of the correct term pound/(inch*inch).
Now this is not the end of your errors in your maxima script. How could you use "modulus" from the Gates document for a glass core? The modulus in the Gates document is for the whole belt not for the core only. If you would want to use it you cannot consider only the glass cross section but use Note 4 of the gates document. In your script, you should have used the modulus for glass fiber only which you specified in your first post:

Quote
A2
Fiberglass
Tensile Strength 350,000 lbs/in^2
Elongation at break 2.5 – 3.5%
Modulus 10,000,000 lbs/in^2

So the final proper term for yung modulus in the maxima script should be (10000000 * pound / (inch*inch)).
Look how it looks when it is correct:

zsh> rmaxima
Maxima 5.32.1 [maxima.sourceforge.net]
using Lisp SBCL 1.1.14
Distributed under the GNU Public License. See the file COPYING.
Dedicated to the memory of William Schelter.
The function bug_report() provides bug reporting information.
(%i1) A: (6.423*pound) * (39.37*inch);
(%o1)                        252.87351 inch pound
(%i2) B: (15 10000000*pound/(inch*inch))*(0.000434*inch*inch);
(%o2)                            651 4340.0 pound
(%i3) A/B;
(%o3)                       .03884385714285714 .05826578571428571 inch
(%i4) ^D
zsh>

Notice also that I got the result in a nice unit which is inch. It makes sense to measure elongation in inches. Notice you got the resutl in 1/inch (your inch is in denominator). How could you measure elongation (i.e. length) in (1/inch)?

Quote
A2
You are confusing wxMaxima place holder (*) as multiplication.

No, I'm not. * means multiplication in maxima. It is no special kind of placeholder. It is the sign for multiplication in maxima. So you did not show my proof of you being wrong to be incorrect.

Quote
A2
hercek, you continue to claim that the Gates equation is wrong, then why don't you show us the correct equation?

Do not make things up! I did not say the Gates equation is wrong. I saied there is a contradiction in the document.

Quote
hercek
Hell, that documents must be writen by heretics of physics. Not only the authors use "hogshead" units, but they cannot even get it right, even with them. Ok, so they define modulus as lb/in² but (based on the formula (in note 7) for ussage of their modulus values) the unit is actually only lb. Or at least I hope this is what they intended (one cannot be completely sure when they have contradictions even in such a simple document).

I said that either their formula is wrong or the unit for their special kind of modulus is wrong. Here I even posted a proof that their document contains a contradiction: [forums.reprap.org]

Edited 1 time(s). Last edit at 02/08/2014 10:42AM by hercek.

As I mentioned before I thought that maxima was using (*) as a place holder,
based on an example script that you hercek showed.

Why didn't you pick up on this, and point it out?

Quote
A2
Why didn't you pick up on this, and point it out?

Sorry. I did not even know nophead joined the discussion at the time I was writing my last response. Once I started to write it I did not see the thread progress. However I'm glad nophead was better at explaining what is wrong. I was trying to do it in a more formal way (without actually modeling it in a proof assistant like COQ or Isabelle) to build a more precise argument showing where the errors are (so that they can be understood and corrected). Nophead stated how it should be. I would not accept it myself if it would not already map perfectly at what I believed it worked like. It would be merely another hypothesis to me. The question which hypothesis is the right one would still be open.

sorry for interrupt!!!

some people use T5 timing belt and some use T2.5. Does it requires any change in firmware? waiting for your reply!