Scaling up Kossel Mini

Quote
hercek
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.

Quote
A2
@ A2

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
]

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.

Quote
nitewing76

Quote
hercek
Hell, that documents must be written 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 usage 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).

hercek...as I constantly have to tell my 11yr old daughter, "It is ok to be wrong, but it is NOT ok to assume you are correct." And there are no "hogshead" units in that document.

Quote
nitewing76
Though it should read, "((lb/in²)/in)" or (psi/in). Remember...when there is no value in front of a unit of measure, it is understood to have a value of one. Therefore if his GT2 belts were 2" wide, then the value would have been 36,000 psi/2in.This is a unique case where the units of measure do not cancel because the author must show the derivative value of ^E to avoid confusion.

1. In the expression lb/in²/in there is no cancelling (even if you would want to cancel anything) because the only way you can simplify that is by writing it as lb/in³.
2. Imperial users like to specify Young's modulus in psi (force per square inch of material cross-section area). Since the Gates document deals only with belts of specified profiles and variable width then it makes sense to use unit lbf/in (force per inch width of a belt). That is because the belt profile height is fixed. Or if we consider only one specific belt then the unit can be lbf only (foce for given belt). But hardly it makes sense to define it as force per cubic inch ... which is what you proposed.
3. Notice you used unit lb when talking about Young's modulus above. In the Young's modulus, the lb actually stands for force (not for mass). You should have used lbf if you want me to even consider first half of your later message (where you complain about that) as anything else than an attempt to confuse me. Finally attention to detail is what you preach, right?

Quote
nitewing76

Quote
hercek
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.

The author gives an equation of BE = ((BSL) x (TL)) / TM. Your equation above is BSL = TM / ((BSL) x (TL). Why did you use the reciprocal of the author's equation? ATTENTION TO DETAIL!!

• First problem with that claim is that you do not have parentheses balanced there so I'm not sure what you mean.
• The second problem is that I think I did not use it. Can you cite me precisely where I used it. Leave out the text around, cite only my symbolic equations or my computations with numbers.

Quote
nitewing76

Quote
hercek
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?

You really need to stop assuming you are correct all the time and conduct some research before your next post. Please, take the time to verify your thoughts as being accurate. You're "assumptions" are the cause of your bad math.

BeltElongation = in...Yes
BeltLength = in...Yes
TensileLoad = lb...NO NO NO! TensileLoad = Force (F) = lbf* (in*s²) or N or kg*(m/s²) The Meaning of Force.

Quote
nitewing76
One time a professor said, "Never calculate Force in anything other than SI units, because it's too easy to screw-up the cancelations."

Quote
nitewing76

Quote
hercek
As for as the rest of your [A2's] 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.

Did you ever stop to think the reason why the belt elongation being cubic centimeters seemed wrong to you was because your math is wrong?