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

kiffe 59877 18
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How do I convert a 1.69 Nm bearing preload torque into the lever-arm length for a 2.5 kg weight in a homemade measuring tool?

A 2.5 kg load corresponds to about 24.5 N of force, so to get 1.69 Nm you need a lever arm of about 0.069 m, i.e. 6.9 cm from the center; using g≈10 m/s² gives about 6.76 cm [#17468831][#17468897] The earlier 7.29 cm result came from mixing kilograms and newtons, because 2.5 kg is not 2.5 N but roughly 24.5 N [#17468831] If you prefer the rough workshop rule, 1 Nm is about 100 g at 1 m, so 1.69 Nm is about 1.7 kg at 10 cm [#17462711][#17468864] The key relation is torque = force × arm length, so once the force is known, the arm length follows directly [#17468831]
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  • #1 17462606
    kiffe
    Level 10  
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    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
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  • #2 17462686
    ociz
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    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.
  • #3 17462711
    kortyleski
    Level 43  
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    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.
  • #4 17462712
    kiffe
    Level 10  
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    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.
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  • #5 17462803
    ociz
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    If you have an expansion sleeve, the "preload" does not matter at all, because the shaft will be loose after that.
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    #6 17462822
    kortyleski
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    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.
  • #7 17463296
    kiffe
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    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 ...
  • #8 17467579
    Rafikusa
    Level 18  
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    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
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  • #9 17468495
    kiffe
    Level 10  
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    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.
  • #10 17468670
    invisibleman
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    Apply a symmetrical arm - identical on both sides of the axis, you will eliminate the influence of arm mass.
  • #11 17468687
    kiffe
    Level 10  
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    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.
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  • #12 17468758
    Rafikusa
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    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.
  • #13 17468801
    kiffe
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    Earthly attraction will eliminate by measuring in a horizontal plane. So how do you do the calculations in this situation?
  • #14 17468831
    Rafikusa
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    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
  • #15 17468844
    kiffe
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    I came out 7.29cm, 6.9cm you, and Kortleski 1.69kg per 10cm. Three different results ... and who is wrong here?
  • #16 17468864
    Rafikusa
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    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?
  • #17 17468874
    kiffe
    Level 10  
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    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?
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    #18 17468897
    Rafikusa
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    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
  • #19 17468908
    kiffe
    Level 10  
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    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!
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Topic summary

✨ The discussion revolves around converting a torque specification of 1.69Nm for preloading tapered roller bearings in a differential repair. The author seeks a method to measure this torque without a precision torque wrench, proposing a makeshift tool using a flat bar and a known weight. Various responses clarify the relationship between torque, force, and lever arm length, emphasizing the need for accurate calculations. The consensus suggests that applying a force of 2.5 kg at a specific distance can achieve the required torque, with calculations indicating that a lever arm of approximately 6.76 cm is necessary. The conversation highlights the importance of considering the mass of the arm and the gravitational force in the calculations.
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FAQ

TL;DR: For 1.69 Nm pinion preload, use 2.5 kg at 6.9 cm or 0.25 kg at 69 cm; "the arm would be 1.69 Nm/24.5 N = 0.069 m." [Elektroda, Rafikusa, post #17468831]

Why it matters: This helps DIY car owners set tapered-roller bearing preload accurately without an inch‑pound torque wrench.

Quick Facts

How do I convert 1.69 Nm preload into a weight and lever length?

Use torque = force × radius and force = mass × 9.81 m/s². For 1.69 Nm, 2.5 kg at 0.069 m works because 2.5 × 9.81 × 0.069 ≈ 1.69. Alternatives: 0.5 kg at 0.345 m, or 0.25 kg at 0.69 m. These give the same torque while reducing sensitivity to arm mass and scale limits. This approach lets you set rotating torque without a dedicated inch‑pound wrench. [Elektroda, Rafikusa, post #17468831]

What arm length should I use if I can pull with 2.5 kg?

Compute radius r = 1.69 Nm ÷ (2.5 kg × 9.81 m/s²) = 0.069 m. That is 6.9 cm from the rotation center. Keep the arm horizontal and light to minimize errors. This compact setup fits inside most pinion flanges and allows steady readings on a small kitchen scale. [Elektroda, Rafikusa, post #17468831]

Is 1.72 kg at a 10 cm arm correct for 1.69 Nm?

Yes. 1.69 Nm ÷ 0.10 m = 16.9 N. Convert to kilograms: 16.9 N ÷ 9.81 m/s² = 1.72 kg. That means a scale reading of 1.72 kg at a 10 cm radius matches the target rotating torque. It’s a convenient midrange option when your scale resolution is best around 1–2 kg. [Elektroda, Rafikusa, post #17468864]

Can I measure pinion bearing preload without a specialized torque wrench?

Yes—use a kitchen scale and a light arm. How-To:
  1. Attach a rigid, lightweight spoke or bar to the flange, horizontal.
  2. Rotate the scale 90° and tare it to zero at your chosen radius.
  3. Pull steadily until the scale shows the target kg equivalent. This substitutes for an inch‑pound wrench for rotating torque checks. [Elektroda, kiffe, post #17468687]

Does the arm’s weight affect the reading, and how do I cancel it?

Arm weight can bias readings, especially at low torques. Use a symmetrical arm with equal mass on both sides of the axis. That cancels the gravitational effects as you rotate and keeps the indicated force focused on bearing drag. It’s a simple way to improve accuracy when using makeshift tools. [Elektroda, invisibleman, post #17468670]

Do I need to worry about gravity if I push horizontally on a scale?

If the scale reads 2.5 kg, you are applying about 24.5 N, regardless of orientation. As one expert put it, “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.” Focus on steady pull and correct radius. [Elektroda, Rafikusa, post #17468831]

What’s the difference between nut torque and bearing rotational torque?

Nut torque crushes the sleeve or seats the bearings. Rotational torque is the drag measured while turning the assembled shaft. You set the nut, then verify rotating torque with your scale-and-arm fixture. “This is not just the tightening torque but the tightening torque of the bearings.” Measure rotation, not only the nut. [Elektroda, kortyleski, post #17462711]

What preload range should I expect for new vs worn bearings?

Manufacturers specify different rotating torque values for new versus used bearings. The thread cites a common example of about 12–23 in‑lb for new bearings. Always use the service manual value for your axle model and bearing state to avoid noise or overheating. [Elektroda, kiffe, post #17462712]

How do crush sleeves change the setup process?

Crush sleeves require significant nut torque to collapse to the correct distance. Then confirm by measuring the rotating torque of the pinion assembly. As noted, “the determinant of this proper dimension is the force needed to rotate such a twisted assembly.” A pipe on the wrench may be needed, e.g., Mercedes W107 hubs. [Elektroda, kortyleski, post #17462822]

What if my pinion threads or nut are not perfect?

Edge-case warning: damaged threads or burrs add friction and can mislead torque feel. With such small torques, any defect risks under-preload or variability. Replace questionable fasteners and sleeves before setting preload to protect new bearings from early failure. [Elektroda, ociz, post #17462686]

My kitchen scale tops at 3 kg. Which combinations work for 1.69 Nm?

Use these equivalents: 2.5 kg at 6.9 cm; 0.5 kg at 34.5 cm; 0.25 kg at 69 cm. All produce about 1.69 Nm. Pick the combo that fits your workspace and keeps the scale within its best measurement range. Longer arms with lighter weights often give smoother control. [Elektroda, Rafikusa, post #17468831]

Do I need to tare the scale when changing its orientation?

Yes. If you rotate the scale 90°, tare it to zero again. Orientation changes how the plate’s weight loads the sensor. Taring ensures the display shows only your applied pull, not the apparatus weight. This makes horizontal pulling measurements consistent and repeatable. [Elektroda, kiffe, post #17468687]

Any tip to minimize measurement error at such low torques?

Use a short 0.1 m arm and increase the weight proportionally to reach target torque. A lighter arm reduces inertia and balance issues. Keep the arm rigid and the pull smooth. Recheck readings after minor adjustments to the pinion nut. [Elektroda, kortyleski, post #17462711]

Why doesn’t 10 N equal exactly 1 kg?

Force depends on gravity. Use F = m × g with g ≈ 9.81 m/s². Example: 2.5 kg corresponds to 24.5 N, not 10 N. When converting between newtons and kilograms on a scale, always multiply or divide by 9.81 to stay accurate. [Elektroda, Rafikusa, post #17468831]
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