Spoke Tension Formula...

[TheDarkInstall]

Alright,

I am in the process of coding up an all encompassing wheel data system in Excel, and so far I have the spoke length, spoke angle, and angle l/r ratio elements done, but I wanted to add in a feature which can predict non-drive side tension based on drive side tension, spoke dimensions, spoke length, angle and hub dimensions...

Has anyone managed to code this before (ie., is it actually possible to accurately create a formula for?), and if so would you mind sharing any insight on how you did this, and what challenges you found along the way, regarding stuff like material of spoke, etc.

Cheers.

[DutchMountains]

Can't help you with a formula, but the spoke calculator at leonard.io gives "Tension ratio L/R" when you click on the "Show complete wheel data" button. It's all javascript, so you should be able to reverse engineer the formula. Or maybe just send Leonard an e-mail asking for the formula's he is using

[kulivontot]

I think it's pretty simple. It's a ratio of the spoke bracing angles.

[TheDarkInstall]

Can't help you with a formula, but the spoke calculator at leonard.io gives "Tension ratio L/R" when you click on the "Show complete wheel data" button. It's all javascript, so you should be able to reverse engineer the formula. Or maybe just send Leonard an e-mail asking for the formula's he is using

Thanks. I have contacted him.

I think it's pretty simple. It's a ratio of the spoke bracing angles.

Not sure it is...

Example;

Tune Mag170 24h, 568mmERD rim

DS angle = 3.81°
NDS angle = 7.70°

Total=11.51°

This gives an angle distribution of DS - 33.1%, NDS - 66.9%

NDS is 33.8% more than DS, so applying this to tension, assuming DS is 100% tension, NDS would be 100 - 33.8 = 66.2% of the DS tension.

Applying this to real world numbers would be, for example;

DS; 115kg
NDS; 76.13kg

A Tune Mag170 does not get anywhere near 76kg on the NDS...

I realise I might be going wrong somewhere with my calculations, but this is what I am seeing...

Any insight would be most welcome!

[kafreeman]

It's not the angle divided by the sum, it's just the angle. NDS/DS

Look for a simple pulley rope tension example in physics.

Sent from my SM-G935V using Tapatalk

[TheDarkInstall]

It's not the angle divided by the sum, it's just the angle. NDS/DS

Look for a simple pulley rope tension example in physics.

Sent from my SM-G935V using Tapatalk

Thanks a lot.

[ergott]

Damon Rinard did a spreadsheet a long time ago that has the calc you're looking for.

http://www.sheldonbrown.com/rinard/spocalc.htm

[Kurets]

In essence it is really simple. The spoke is pulling on the rim diagonally and the tensioned spoke can be considered a force vector. The force vector in a cross section of the wheel can be divided into two components, the radial and the lateral component. The radial component is controlled by the cosine of the bracing angle and the lateral by the sine of the bracing angle.

Your example:
Tune Mag170 24h, 568mmERD rim

DS angle = 3.81°
NDS angle = 7.70°

Will give that the lateral component of spoke tension on the DS is 0.065 and 0.135 on the NDS. If you want to center the wheel both must be equal which means that the tension on the NDS must be 0.065/0.135 = 0.48 times the tension on the DS. This would give (if 115kg on the DS) a force of 55kg in the NDS spokes which sounds about reasonable.

So tension in a NDS spoke should be: Tension(DS)*sin(angle(NDS))/sin(angle(DS))

Please correct me if I am wrong, this is just how I feel intuitively that it should work and the numbers seem ok-ish.

[kafreeman]

That's correct, a spoke is a two force member.

The force component in each direction is found by the sin / cos of the angle. Primary contribution in radial forces will be the DS. NDS is there to balance the lateral but will contribute much less in the radial direction.

I've only built a couple wheels, the first one I got up to tension but was way out radially. Second wheel I got the spokes with enough tension not to wiggle around then put a dial indicator on it and brought tension up slowly. Came out pretty good much faster than the other wheel.

Sent from my SM-G935V using Tapatalk

[TheDarkInstall]

Wow, thanks a lot for the replies. Good info!

[TheDarkInstall]

Tension(DS)*sin(angle(NDS))/sin(angle(DS))

Please correct me if I am wrong

Thanks for your detailed reply. Makes sense.

You got NDS and DS the wrong way round though

[Kurets]

Yeah, that always happens. In the end it is easy, if your answer ends up wrong you flipped the division somewhere.
The equation:
tension(DS)*sind(angle(DS))=tension(NDS)*sind(angle(NDS))
Is a better choice to derive the tensions from. Sin increases around 0 so an increase in bracing angle will lead to a reduction in spoke tension.

Sent from mTalk

[TheDarkInstall]

In essence it is really simple.

I am clearly a bit of a trigonometry noob, so excuse my naivety if these are obvious questions, but...

tension(DS)*sind(angle(DS))=tension(NDS)*sind(angle(NDS))

What is the difference here between using sin() and sind()?

Also, how would NDS tension be derived, if only the DS is known? In the case of the example, would this be;

115*sind(0.065)=x*sind(0.135) ?

And going back to your previous post, some even more basic questions...

The radial component is controlled by the cosine of the bracing angle and the lateral by the sine of the bracing angle.

Why is this the case?

The lateral component of spoke tension on the DS is 0.065 and 0.135 on the NDS.

I realise these are the DS and NDS bracing angles converted to Radians, but can't get my head around how this equates to the lateral (Fx) element of the Force Vector... Any chance you can explain how this works? Again, this is probably really obvious to those with got a decent mathematics education in school...

So tension in a NDS spoke should be: Tension(DS)*sin(angle(NDS))/sin(angle(DS))

Same question again then... why is the sin of the angle used here... ?

This is basically me asking you to explain trigonometry to me, I know...

[Kurets]

First I suggest you should look up a basic textbook on trigonomertry. But I will answer your questions as good as I can.

What is the difference here between using sin() and sind()?

I write sind to clarify that I mean sin in degrees and not radians.

Also, how would NDS tension be derived, if only the DS is known? In the case of the example, would this be;
115*sind(0.065)=x*sind(0.135) ?

No, the result sind(3.81) is 0.065 (and sind(7.7) is 0.135). What it means is that (on the DS) a force pulling in the direction of the spoke will pull the rim laterally with 0.065 of the total force. This comes from that you can describe any vector in a plane by either its x and y components or its angle and magnitude. For the spoke the interesting angle is the bracing angle and the magnitude is the tension in the spoke.
Note that the radial component is very close to 1 (for the DS 0.998 and for the NDS 0.991 ) which could make you think that the sum is greater than the total tension in the spoke. If that is the case, look up the Pythagorean theorem.

Same question again then... why is the sin of the angle used here... ?
This is basically me asking you to explain trigonometry to me, I know...

Look up the unit circle and what cosd(angle) and sind(angle) mean. You can look at it like the shadow the spoke would cast if it was illuminated perpendicularly to the chosen reference plane. I think this is all tricky to explain in a forum, and sitting down with some paper would be better. But, all in all, you should look at the definitions of the trigonometric expressions and the basics of trigonometry.
Maybe I can post some pictures later if you want, but really it is basic stuff.

EDIT: I can dig up some links and PM you later tonight. Mainly wiki. But the maths pages are always good on wikipedia.

[Kurets]

Ok, so I realize I cannot send PMs as I guess I don't have enough posts yet.

To understand basic trigonometry it is necessary to understand the unit circle: https://en.wikipedia.org/wiki/Unit_circle it is a very simple illustration of how an angle and a magnitude (in that case 1) can describe a unique point in a plane. And, that sin describes the projection on the "y-axis" and that cos describes the projection on the "x-axis". (I write with quotes as there are no general y and x axes, but it is common to call them by those names)

From there you can undestand that cos(0) = cosd(0) = 1 and that sin(0) = sind( 0) = 0 which tells you that any vector which has an angle of 0 to your chosen reference line (i.e. it is parallel to the reference line) has no component perpendicular to the reference line. For a spoked wheel, that would mean that a bracing angle of 0 will give a wheel that has no lateral stiffness arising from the tension in the spokes.

By increasing the angle to 45degress you get that cosd(45) = sind(45) = 1/sqrt(2), both components are identical! This is intuitively true as well, the 45 degree angle is 45 degrees from the x-axis and 45 degrees from the y-axis so the projection will be the same on both axes. For a spoked wheel it would mean that the wheel would be very stiff laterally, but excessively so, and it would be a very wide wheel, unless it had an impractically small diameter.

So, once you understand the concept of projection onto the reference axes it becomes easy. The small bracing angle means that the main force acting on the rim is pulling it inward toward the hub, radially(so it must be the cosine of the bracing angle). The opposite is true for the lateral component, therefore it must be the sine of the bracing angle.