Converting 1.69Nm Tension to KG for Differential Repair Using Tapered Roller Bearings

Hello,
seemingly a simple task. But from the beginning. I have to repair the differential in my old car. I need to replace the tapered roller bearings and preload them. The instruction states that after installing new bearings (two tapered roller bearings on the attacking shaft and tightened with a screw) the initial tension is to be 15 InLbs (let's assume it is about 1.69Nm). To measure it you need to have a torsion wrench (or a mega precise clock torque wrench). Unfortunately, I do not have one. Another thing is that they are very expensive and it does not pay to buy one job for myself. I wanted to make a simple tool: a profile / flat bar of a certain length, which I will put on the flange in a differential and at a specific point I will apply a specific force. I assume 2.5 kg will be "strength", but I don't know how long I should put it on. I've made some calculations, but I'm not sure about them.
Well, 10N is 1 kg. So 1.69Nm is 0.169kg / m. (0.338kg at 0.5m, 0.676 at 0.25m, 1.352kg at 0.125m etc). Converting this further it turned out to me that 2.5kg should be applied at a distance of 7.29cm from the center of the axis of rotation.
Is the calculation / deduction correct?
I would add that using online calculators, 15InLbs converts to approx. 1.69N / m and converts to 0.1714kg / m or to 0.172kg / m. Calculating in different ways does not come out the same value.
Thank you for your help

Such small moments can be taken on new elements, without the slightest flaw on the thread, because any fern will cause you to tighten too weakly.

This is not just the tightening torque but the tightening torque of the bearings.
Roughly it can be assumed that 1Nm is 100 grams on a horizontal arm with a length of 1m. but you will need to take into account the weight of the arm ... Therefore it will be safer to take a 0.1m arm and increase the weight proportionally to 1kg.
Here you need 1.7Nm or 1.7 kg per 10cm light arm.

Bearings with raceways as well as expansion sleeves and simmering will be new. The attack shaft and nut remain old. The manufacturer specifies different values for old bearings and different values for new bearings. Of course, these values are given in the range e.g. 12-23 InLbs. Each service gives information on the need to preload the bearings. Turning such a twisted attack shaft, the torsion key should be 1.69Nm. In other words, such a twisted attack shaft should create initial resistance.
Well, but is my calculation correct? The values are very small, so the error field is also small.

If you have an expansion sleeve, the "preload" does not matter at all, because the shaft will be loose after that.

ociz wrote:

If you have an expansion sleeve, the "preload" does not matter at all, because the shaft will be loose after that.

Have you ever done anything like this? To crush this sleeve to size, you need to tighten securely. And the determinant of this proper dimension is the force needed to rotate such a twisted assembly. In mercu w107 when screwing a new rear hub (identical bearing design), a pipe meter must be put on the key and tug well to achieve the correct preload.

So my calculation is wrong? So when applying 2.5 kg, what should the arm length be? I will add that the measurement will be done in a horizontal plane to eliminate the weight of the arm (and precisely writing so that the frame in each angular arrangement has the same mass). Does my way of obtaining bearing preload make sense? This is due to the lack of access to such a precise torque wrench ...

The 2.5kg pull force under gravity is:
2.5 kg × 9.81 kg / m? = 3.03 N.
To obtain a torque of 1.69Nm, a force of 3.03 N must be applied to the arm 1.69 Nm / 3.03 N = 0.56 m

Well, the calculation is rather wrong. First, it does not take into account the mass of the arm, which in different angular positions "changes the mass of inertia". Secondly, to make the measurement uniform in the full 360 * range, I will perform it in a horizontal position, i.e. I eliminate the variable weight of the arm.

Apply a symmetrical arm - identical on both sides of the axis, you will eliminate the influence of arm mass.

That's also true ... will use an electronic (kitchen) scale for measuring. The measuring range is up to 3kg. Of course, after turning the weight by 90 * I have to tare it, because the weight of the plate will decrease. Instead of a shoulder, he will use a thick spokes: rigid and minimal weight. But I still don't know what the length of the arm must be .... there are divergent opinions about this.

And what is wrong?
You start at 2.5 kg. The only sensible combination of mass and strength is the force with which the earth attracts this mass.
How do you think kitchen scales work? They measure the force with which the plate is pressed against the rest of the weight and, assuming that the weight is on a horizontal surface, this force is converted into mass.
Just like in elementary school: F = m × a. "F" is force, "m" is mass, "a" is acceleration (in this case gravitational acceleration).
If your kitchen scale shows 2.5 kg, it means that you press the plate on the base with 3.03N (after taring).
If you want to achieve a torque of 2.69Nm, you put the weight on the 0.56m arm and press until the weight shows 2.5kg.

There can be no discrepancy in terms of arm length with your data.
With such a task, it is assumed that the arm is attached so that its mass does not introduce an error. This must be done by the executing person.

Earthly attraction will eliminate by measuring in a horizontal plane. So how do you do the calculations in this situation?

Exactly as I showed.
But now I can see that at the beginning I hit a bull and then I didn't get it anymore.
2.5 kg × 9.81 kg / m? is 24.5N, not 3.03 (I have no idea where it came from).
the arm would be 1.69 Nm / 24.5 N = 0.069m, so completely not practical.

Imagine a hanging spring balance. If you hang 2.5 kg on it, it means that the mass pulls with a force of 24.5 N. And now it does not matter how and where you pull this weight, if it shows 2.5 kg, then you pull with a force of 24.5 N.
It would be better to choose 250g, because then the frame is 0.69m. It can also be 500g and 0.345m

I came out 7.29cm, 6.9cm you, and Kortleski 1.69kg per 10cm. Three different results ... and who is wrong here?

1.69kg per 10cm is rounded. That's exactly 1.72kg, because it comes out 16.9N per 10cm.

1.69Nm / 0.1m = 16.9N
that is, the weight must show 16.9N / 9.81 = 1.72kg

How did you get this 7.29cm?

In the first post I gave my calculation method.

Added after 12 [minutes]:

My result is 2.5 kg. That is 2.5 kg applied to a 7.29 cm lever. You came out 1.72kg out of 10cm. Kortyleski calculated 1.69kg per 10cm. So only my result deviates from your calculations ... This is now a trivial question: what arm length at 2.5kg?

As I wrote after the correction:
1.69 Nm / 24.5 N = 0.069m.
Or with rounding at G = 10m / s?: 1.69N / (2.5kg × 10 m / s? = 0.0676m

OK, so coming to the end of these calculations ... 2.5 kg applied to a 6.76 cm long arm from the center of rotation. So now I would ask you to check at what stage I made a mistake?
Well, 10N is 1 kg. So 1.69Nm is 0.169kg / m. (0.338kg at 0.5m, 0.676 at 0.25m, 1.352kg at 0.125m etc). Converting this further it turned out to me that 2.5kg should be applied at a distance of 7.29cm from the center of the axis of rotation.

Added after 6 [minutes]:

Okay, I made a math error. Everything is correct! 0.169kg per 100cm (1 meter), the same as 2.5kg per 6.75cm. Thank you for your help, I'm closing the topic!