How to Calculate Load on a Linear Guide

Oct 7

To calculate load on a linear guide, determine every force acting on the moving assembly, locate where each force acts relative to the rails and carriages, calculate the resulting pitch, yaw, and roll moments, and then use static-equilibrium relationships to determine the load carried by each individual carriage.

The most important point is:

Do not divide the total machine load by the number of linear guide blocks unless the load is perfectly centered and the geometry truly produces equal load sharing.

Real machines often have:

offset payloads,
high centers of gravity,
acceleration and deceleration,
cutting or process forces,
unequal rail spacing,
unequal block spacing,
vertical or wall mounting.

Those conditions can make one carriage carry substantially more load than the others.

THK's current linear-guide selection process specifically requires designers to calculate the applied load on each LM block, then convert multi-directional loading into equivalent load where necessary before checking static safety and fatigue life.

Key Takeaways

Calculate load on each carriage, not just the total guide system.
Start with weight: W = m × g.
Include acceleration forces using F = m × a.
Include process forces such as cutting, pressing, clamping, or belt forces.
Calculate moment loads using M = F × d.
Evaluate pitch, yaw, and roll separately.
Rail spacing helps resist roll moment.
Block spacing helps resist pitch and yaw moments.
An off-center center of gravity can heavily load one carriage even when total weight is modest.
Vertical and wall-mounted axes require different force directions than horizontal systems.
Simultaneous radial and lateral loads may require an equivalent-load calculation.
Peak carriage load is used for static-safety evaluation.
Varying loads over the machine cycle should be converted to an appropriate fatigue-equivalent average load for life calculations.
Do not assume four blocks each carry exactly 25% of the load.
Manufacturer selection equations should control the final design.

The Basic Linear Guide Load Calculation

For a simple horizontal system with a centered payload and no acceleration:

W = m × g

where:

W = weight force in newtons,
m = moving mass in kilograms,
g = gravitational acceleration, approximately 9.81 m/s².

If four identical guide blocks genuinely share the load equally:

P = W / 4

where:

P = load per block.

Simple Example

Moving mass:

100 kg

Weight:

W = 100 × 9.81

W = 981 N

With four equally loaded blocks:

P = 981 / 4

P = 245.25 N per carriage

That is the easiest possible case.

Real industrial applications usually require more analysis.

Why Total Weight Is Not Enough

Suppose the same 100 kg load is positioned much closer to one end of the guide table.

The total weight is still:

981 N

But the carriage reactions are no longer equal.

Blocks near the load may carry much more than:

245 N

while blocks farther away carry less.

The linear-guide system responds not only to force but also to:

where the force acts.

That introduces moments.

Step 1: Identify Every Applied Force

Begin with a free-body diagram of the moving assembly.

Include:

payload weight,
moving-table weight,
fixture weight,
motor or gearbox weight carried by the axis,
cable-chain forces where significant,
acceleration forces,
deceleration forces,
cutting forces,
pressing forces,
belt tension,
pneumatic or hydraulic forces,
external machine reactions.

THK's applied-load guidance specifically identifies mounting orientation, center of gravity, acceleration/deceleration, external forces, and guide arrangement as factors that must be considered when calculating carriage load.

Step 2: Calculate Weight

For every moving mass:

W = m × g

Example:

Moving assembly:

250 kg

Then:

W = 250 × 9.81

W = 2,452.5 N

That force acts vertically downward through the assembly's center of gravity.

Step 3: Locate the Center of Gravity

Record the center of gravity relative to the linear guide system.

You need dimensions such as:

distance between rails,
distance between blocks,
horizontal offset from rail centerline,
vertical height above the guide plane,
longitudinal offset along the direction of travel.

The farther the center of gravity is from the guide support footprint, the larger the resulting moment can become.

Step 4: Calculate Moment Loads

Moment is:

M = F × d

where:

M = moment,
F = force,
d = perpendicular distance from the force to the reference point or support plane.

For linear guides, moments are typically discussed as:

pitch,
yaw,
roll.

Pitch Moment

Pitch rotates the moving structure forward or backward.

A common cause is a load positioned:

ahead of,
behind,
above

the carriage group.

If an acceleration force acts at a height above the rails, it can create pitch.

Example

Acceleration force:

1,500 N

Center of gravity:

0.20 m above the guide plane

Pitch moment:

M = 1,500 × 0.20

M = 300 N·m

That moment must be resisted by the carriage arrangement.

Roll Moment

Roll rotates the table from side to side.

A lateral force acting above the guide plane is a common cause.

Suppose:

lateral force = 2,000 N,
height above rail plane = 0.30 m.

Then:

Mroll = 2,000 × 0.30

Mroll = 600 N·m

A two-rail system resists that moment largely through different vertical reactions at the two rails.

Yaw Moment

Yaw rotates the table horizontally.

For example, a lateral force applied:

ahead of the carriage group,
behind the carriage group

can create yaw.

If:

force = 1,000 N,
longitudinal offset = 0.40 m,

then:

Myaw = 1,000 × 0.40

Myaw = 400 N·m

How Rail Spacing Affects Load

For a simplified two-rail system resisting roll moment:

ΔF ≈ M / S

where:

ΔF = reaction-force couple,
M = applied roll moment,
S = distance between rails.

Example

Roll moment:

600 N·m

Rail spacing:

0.50 m

Then:

ΔF = 600 / 0.50

ΔF = 1,200 N

One rail's reaction increases while the opposite rail's reaction decreases.

This is why wider rail spacing can substantially reduce individual carriage loading caused by roll.

Wider Rails Mean More Mechanical Leverage

Take the same:

600 N·m

roll moment.

Rails 300 mm Apart

ΔF = 600 / 0.30 = 2,000 N

Rails 600 mm Apart

ΔF = 600 / 0.60 = 1,000 N

Doubling rail spacing cuts the required reaction-force couple in half in this simplified example.

That is often more effective than simply choosing a larger guide.

How Block Spacing Affects Moment Load

Two blocks on one rail can resist pitch or yaw through their longitudinal separation.

A simplified relationship is:

ΔF ≈ M / L

where:

L = spacing between block reaction centers.

Example

Pitch moment:

300 N·m

Block spacing:

0.20 m

Then:

ΔF = 300 / 0.20

ΔF = 1,500 N

If block spacing increases to:

0.40 m

then:

ΔF = 300 / 0.40

ΔF = 750 N

Again:

better geometry reduces carriage reaction loads.

Calculating Load on Four Linear Guide Blocks

A very common arrangement is:

two parallel rails,
two blocks on each rail.

Label them:

P1 = front-left,
P2 = rear-left,
P3 = front-right,
P4 = rear-right.

For a perfectly centered static vertical load:

P1 = P2 = P3 = P4 = W / 4

But if the load is offset, the carriage reactions must satisfy:

Force equilibrium

P1 + P2 + P3 + P4 = W

and:

Moment equilibrium

The carriage reactions must also balance the applied:

pitch,
yaw,
roll moments.

This is why simple division is often insufficient.

A Simplified Four-Block Example

Assume:

mass = 200 kg,
total weight = 1,962 N,
rail spacing = 0.40 m,
block spacing = 0.60 m,
payload centered longitudinally,
payload offset laterally enough to create 200 N·m roll moment.

Base static load per block:

1,962 / 4 = 490.5 N

Roll reaction between rails:

ΔF = M / S

ΔF = 200 / 0.40

ΔF = 500 N

Half of that rail-level difference is distributed across the two blocks on each rail in this simplified symmetric case.

Additional load per block on the heavily loaded rail:

500 / 2 = 250 N

So heavily loaded side:

490.5 + 250 = 740.5 N per block

Light side:

490.5 − 250 = 240.5 N per block

The total payload never changed.

But one side now carries more than three times the carriage load of the other side.

Why “Divide by Four” Can Be Dangerous

If you had assumed:

490.5 N per block

you would underestimate the most heavily loaded carriage:

740.5 N vs. 490.5 N

That affects:

static safety,
fatigue life,
rigidity,
guide size.

Schaeffler's linear-guide dimensioning guidance likewise emphasizes determining the most heavily loaded carriage from the actual forces and carriage geometry rather than simply dividing system load by carriage count.

Step 5: Add Acceleration Forces

When the machine accelerates:

F = m × a

Suppose:

moving mass = 200 kg,
acceleration = 5 m/s².

Then:

F = 200 × 5

F = 1,000 N

This inertial force acts through the center of gravity.

If the center of gravity is above the guide plane, it also produces a moment.

Acceleration Example With Center-of-Gravity Height

Using:

inertial force = 1,000 N,
CG height = 0.25 m.

Moment:

M = 1,000 × 0.25

M = 250 N·m

That moment changes carriage loads during acceleration.

When the machine decelerates, the force reverses.

Therefore, the most heavily loaded carriage may change during different parts of the motion cycle.

Acceleration Can Matter More Than Payload

Consider a lightweight, high-speed pick-and-place axis.

Mass:

50 kg

Acceleration:

20 m/s²

Inertial force:

F = 50 × 20

F = 1,000 N

Weight is only:

50 × 9.81 = 490.5 N

In this case, dynamic inertial force is roughly twice the gravitational force.

Ignoring acceleration would badly underestimate guide loading.

Step 6: Include Deceleration

Deceleration should be treated as acceleration in the opposite direction.

Fast stops can produce some of the highest guide loads in the machine cycle.

Also consider:

emergency stopping,
collision conditions,
end-of-stroke impact

where relevant to the engineering design.

THK specifically notes that sudden starting and stopping and large moments from overhung loading are reasons a static safety factor must be considered.

Step 7: Add Process Forces

Industrial machinery may apply significant process loads.

Examples include:

machining forces,
drilling forces,
grinding forces,
press forces,
welding reactions,
clamping forces,
web tension,
conveyor belt forces.

These forces must be included in the free-body diagram.

They can act:

vertically,
laterally,
longitudinally,

and often create moments because they act away from the guide plane.

Example: Cutting Force

Suppose a machining process creates:

lateral cutting force = 3,000 N,
tool contact point = 0.20 m above rails.

Moment:

M = 3,000 × 0.20

M = 600 N·m

This moment exists in addition to:

machine-table weight,
workpiece weight,
acceleration loads.

Step 8: Account for Mounting Orientation

Guide loading changes depending on machine orientation.

Common arrangements include:

horizontal,
vertical,
wall mount,
inclined,
inverted.

THK's applied-load guidance provides different calculation cases for:

horizontal mounting,
overhung horizontal mounting,
vertical mounting,
wall mounting,
inclined mounting,
moving rails,
acceleration/deceleration,
external forces.

Horizontal Mount

Gravity generally acts in the radial direction.

This is usually the easiest configuration to visualize.

Vertical Mount

Gravity acts parallel to the direction of travel.

The drive system must support the moving mass against gravity.

Depending on the geometry, the guide carriages may experience:

lateral,
moment,
process loads

rather than simply the full gravitational load in the same way as a horizontal axis.

Wall Mount

Gravity can become a lateral guide load.

Verify the manufacturer's permissible and rated loading in that direction.

Inverted Mount

The guide can experience reverse-radial loading.

Do not assume every guide has identical ratings in:

radial,
reverse-radial,
lateral directions.

Four-Way Equal Load vs. Radial-Type Guides

Some profile rail guides have equal ratings in all four primary load directions.

Others do not.

THK explicitly separates:

four-way equal-load guides,
radial-type guides

and publishes different directional rating factors where required.

Therefore:

load direction matters.

Step 9: Convert Simultaneous Loads to Equivalent Load

A carriage may experience:

radial load,
lateral load

at the same time.

For applicable THK guide systems, equivalent load can be expressed in the general form:

PE = X × PR(L) + Y × PT

where:

PE = equivalent load,
PR(L) = radial or reverse-radial load,
PT = lateral load,
X, Y = manufacturer-specific equivalent-load factors.

THK publishes the applicable factors by guide model.

Do not assume:

PE = PR + PT

for every guide.

Use the manufacturer's equation.

Why Equivalent Load Is Needed

Fatigue-life and static calculations usually require a representative load value.

If the carriage experiences multiple simultaneous directions, the manufacturer provides a method to convert them into a single equivalent value.

That equivalent load is then used in:

static-safety calculations,
fatigue-life calculations.

Step 10: Find the Maximum Carriage Load

Once you calculate:

stationary condition,
acceleration,
constant velocity,
deceleration,
process loading,
maximum payload,

identify the highest load experienced by any one carriage.

Call it:

Pmax

This value is essential for static-safety evaluation.

Step 11: Calculate Static Safety Factor

A common relationship is:

fs = C₀ / Pmax

where:

fs = static safety factor,
C₀ = basic static load rating,
Pmax = maximum applied equivalent load.

THK uses this relationship in its linear-motion selection methodology.

The required safety factor depends on:

vibration,
shock,
operating condition,
manufacturer recommendations.

Do not invent a universal safety factor.

Static Safety Example

Suppose:

C₀ = 30,000 N,
maximum carriage load = 6,000 N.

Then:

fs = 30,000 / 6,000

fs = 5

Whether that is acceptable depends on:

guide type,
machine conditions,
manufacturer guidance.

Why Static Rating Is Not Enough

A guide can safely survive the maximum instantaneous load and still have insufficient fatigue life.

That is why selection requires both:

static-safety evaluation,
dynamic-life calculation.

Step 12: Calculate Loads Over the Full Duty Cycle

Industrial machines rarely operate under one constant load.

A carriage may experience:

high load during acceleration,
lower load during cruise,
high opposite load during deceleration,
different loads during return travel.

For life calculation, these varying loads should be converted into an appropriate fatigue-equivalent average load.

THK defines average load as the constant load that would produce equivalent service life to the varying loads experienced during operation.

Do Not Use a Simple Arithmetic Average

Suppose a guide experiences:

1,000 N for half the travel,
5,000 N for half the travel.

Simple average:

3,000 N

But rolling-bearing fatigue is nonlinear.

The correct fatigue-equivalent average load is not generally the simple arithmetic average.

Use the manufacturer equation for:

balls,
rollers.

THK provides separate calculation methods based on the rolling-element type.

Ball vs. Roller Average-Load Calculation

Because ball and roller fatigue relationships use different exponents, their equivalent average-load calculations differ.

This is another reason not to create a generic formula without first identifying:

ball guide,
roller guide.

Use the exact manufacturer's life method.

Complete Linear Guide Load Calculation Workflow

Use this sequence.

Step 1 — Determine Moving Mass

Include:

table,
payload,
fixture,
tooling,
moving actuator components.

Step 2 — Calculate Gravity Load

W = m × g

Step 3 — Establish Coordinate Directions

Define:

X,
Y,
Z,
radial,
reverse-radial,
lateral.

Step 4 — Locate the Center of Gravity

Record its position relative to:

rails,
blocks,
guide plane.

Step 5 — Record Rail Spacing

Measure center-to-center distance between the rails.

Step 6 — Record Block Spacing

Measure effective center-to-center distance between blocks.

Step 7 — Calculate Inertial Forces

F = m × a

for each acceleration direction.

Step 8 — Record External Forces

Include:

cutting,
clamping,
pressing,
tension.

Step 9 — Calculate Pitch, Yaw, and Roll Moments

M = F × d

Step 10 — Solve Static Equilibrium

Use:

ΣF = 0

and:

ΣM = 0

to determine carriage reactions.

Step 11 — Calculate Each Carriage Load

Determine:

or however many blocks are present.

Step 12 — Evaluate Every Operating Condition

Check:

stationary,
acceleration,
full speed,
process,
deceleration,
emergency conditions where required.

Step 13 — Convert Directional Loads

Use the manufacturer's equivalent-load method.

Step 14 — Find Pmax

Identify the most heavily loaded carriage.

Step 15 — Check Static Safety

fs = C₀ / Pmax

Step 16 — Determine Fatigue-Equivalent Average Load

Use the manufacturer's duty-cycle method.

Step 17 — Calculate Required Life

Compare guide life with machine requirements.

Step 18 — Check Rigidity

Load capacity does not guarantee acceptable deflection.

Worked Example: Two Rails and Four Blocks

Consider a simplified horizontal machine table.

Machine Data

Moving mass:

300 kg

Gravity:

9.81 m/s²

Rail spacing:

500 mm = 0.50 m

Block spacing:

600 mm = 0.60 m

Center of gravity:

centered between front and rear blocks,
100 mm laterally offset,
250 mm above guide plane.

Lateral acceleration:

4 m/s²

1. Calculate Weight

W = 300 × 9.81

W = 2,943 N

Base vertical load with perfect symmetry:

2,943 / 4 = 735.75 N per block

2. Calculate Lateral Acceleration Force

F = m × a

F = 300 × 4

F = 1,200 N

3. Calculate Roll Moment From CG Height

Mroll = 1,200 × 0.25

Mroll = 300 N·m

4. Convert Roll Moment to Rail Reaction

Rail spacing:

0.50 m

ΔF = 300 / 0.50

ΔF = 600 N

With two symmetric blocks on each rail:

600 / 2 = 300 N per block

approximately added to the heavily loaded side and subtracted from the opposite side for this simplified condition.

Heavy-side vertical reaction:

735.75 + 300 = 1,035.75 N per block

Light-side vertical reaction:

735.75 − 300 = 435.75 N per block

The acceleration force also creates lateral carriage loading and therefore must be combined according to the specific guide's directional/equivalent-load method.

The important lesson is:

the guide does not experience only 736 N per block simply because the table has four carriages.

What About the 100 mm Lateral CG Offset?

The static weight itself creates a roll moment:

M = W × d

M = 2,943 × 0.10

M = 294.3 N·m

That static moment should also be included.

So even before acceleration:

the carriages are not equally loaded.

The final calculation combines:

static weight,
CG offset,
acceleration,
directional guide forces.

Center of Gravity Can Matter More Than Mass

Consider two machines with the same:

300 kg payload.

Machine A:

CG 50 mm above guide plane.

Machine B:

CG 500 mm above guide plane.

Under the same lateral acceleration, Machine B produces:

10× the roll moment.

This demonstrates why specifying only:

“300 kg load”

is not enough to size a linear guide.

One Block vs. Multiple Blocks

A single block must directly handle the applied:

forces,
moments.

With multiple blocks, moments can be resisted partly through reaction-force couples created by block spacing.

That can dramatically reduce the moment demand on each individual block.

One Rail With Two Blocks

A two-block single-rail arrangement improves:

pitch resistance,
yaw resistance

through longitudinal block spacing.

But roll moment may still rely heavily on individual carriage moment capacity because there is no second rail creating a wide transverse support footprint.

Two Rails With One Block Each

This provides rail spacing for roll resistance.

But limited longitudinal spacing may reduce the structure's ability to resist:

pitch,
yaw

through block separation.

Two Rails With Two Blocks Each

This common four-block arrangement provides both:

rail spacing,
block spacing.

That is one reason it is widely used for industrial tables.

Why Four Blocks Do Not Guarantee Equal Load Sharing

Even when the theoretical geometry is symmetric, real systems can have unequal loading because of:

mounting-surface error,
table deformation,
block height differences,
preload,
rail parallelism,
structural stiffness.

THK warns that multiple blocks mounted closely can experience nonuniform load distribution due to factors including moment loads and mounting-surface accuracy.

For high-load or precision applications, rigid-body static calculations may therefore be only the first level of analysis.

Flexible Structures Require More Advanced Analysis

Simple statics assumes:

rigid table,
rigid base,
ideal guide positions.

Large machines may require consideration of:

base deflection,
table bending,
rail deformation,
carriage stiffness.

Manufacturer engineering software or finite-element analysis may be appropriate.

Schaeffler's dimensioning approach likewise considers carriage geometry and the actual applied forces to determine the most heavily loaded carriage.

Carriage Load vs. Permissible Moment

Manufacturers may publish:

C,
C₀,
MA,
MB,
MC.

These represent different limits.

C

Basic dynamic load rating.

C₀

Basic static load rating.

MA / MB / MC

Permissible moment values about the manufacturer's specified axes.

Do not compare a calculated moment directly with C or C₀.

They have different units.

Newtons vs. Newton-Meters

Force:

Moment:

N·m

A 5,000 N load is not directly comparable to a:

500 N·m

moment.

Convert the machine geometry into appropriate carriage reactions or compare moments with manufacturer permissible-moment data as required.

Common Linear Guide Load Calculation Mistakes

Mistake 1: Dividing Payload by Block Count

This is only valid under truly symmetric loading.

Mistake 2: Using Kilograms as Force

Mass is in kilograms.

Force should generally be calculated in newtons:

W = m × g

Mistake 3: Ignoring Center of Gravity

Offset loads create moments.

Mistake 4: Ignoring Acceleration

High-speed equipment may generate inertial loads greater than gravity.

Mistake 5: Ignoring Deceleration

Stopping can be a peak-load condition.

Mistake 6: Ignoring Process Forces

Machining, pressing, tension, and clamping forces must be included.

Mistake 7: Ignoring Rail Spacing

Rail spacing strongly affects roll reaction loads.

Mistake 8: Ignoring Block Spacing

Block spacing strongly affects pitch/yaw reactions.

Mistake 9: Assuming Every Guide Has Four-Way Equal Ratings

Directional capacity varies by guide design.

Mistake 10: Using Arithmetic Average Load for Fatigue Life

Rolling-bearing fatigue is nonlinear.

Mistake 11: Checking Only C

Static rating C₀ and static safety must also be evaluated.

Mistake 12: Checking Only Load Capacity

Rigidity and deflection may control the design before load capacity does.

Linear Automation USA's Perspective

At Linear Automation USA, we believe the most important shift in linear-guide sizing is moving from:

“How heavy is the payload?”

to:

“What load does each carriage actually experience?”

Those are not the same question.

A 100 kg payload located:

low,
centered,
between widely spaced rails

can be an easier guide application than a 50 kg payload that is:

heavily cantilevered,
high above the rails,
rapidly accelerated.

We therefore start with:

forces → center of gravity → moments → rail spacing → block spacing → individual carriage reactions

before choosing a nominal rail size.

We also do not use preload or larger bearings as the first solution to poor machine geometry.

If carriage load is high because of a roll moment, increasing:

rail spacing

may reduce loading dramatically.

If pitch load is high, increasing:

block spacing

may be more effective.

If the problem comes from a high center of gravity, lowering the payload can reduce the moment at its source.

Only after the actual carriage loads are understood should the designer compare:

dynamic rating C,
static rating C₀,
permissible moments,
preload,
rigidity,
accuracy,
service life.

Linear Automation USA supports industrial profile rail systems from Schaeffler, SBC Linear, and WON Linear, along with identification, sizing, replacement, and cut-to-length support.

Our rule is:

Never size a linear guide from payload weight alone. Size it from the most heavily loaded carriage under the worst credible operating condition.

Frequently Asked Questions

How Do You Calculate Load on a Linear Guide?

Calculate all applied forces and moments, then use the geometry of the rails and carriages to determine the reaction load on each carriage.

Can I Just Divide the Weight by Four Blocks?

Only if the load is perfectly centered and the guide geometry produces equal load sharing.

How Do I Convert Kilograms to Linear Guide Load?

Use:

W = m × 9.81

to convert mass in kilograms to gravitational force in newtons.

How Does Acceleration Affect Linear Guide Load?

Acceleration creates inertial force:

F = m × a

and can also create moments when the center of gravity is offset from the guide plane.

How Does Rail Spacing Affect Linear Guide Load?

Greater rail spacing reduces the reaction forces required to resist roll moment.

How Does Block Spacing Affect Linear Guide Load?

Greater longitudinal block spacing reduces reaction forces required to resist pitch or yaw moments.

What Is the Most Important Linear Guide Load?

The most heavily loaded carriage under the worst applicable operating condition is especially important for static-safety evaluation.

What Is Equivalent Load?

Equivalent load converts simultaneous directional loads into a single representative load using manufacturer-specific factors.

Should I Use Peak Load or Average Load?

Use peak load for static-safety checks and the appropriate fatigue-equivalent average load for service-life calculations.

Is Average Load Just the Arithmetic Average?

Usually not. Rolling-bearing fatigue is nonlinear, so use the manufacturer's average-load equation.

Does Preload Count as Load?

Preload creates internal rolling-element load and may need to be considered in fatigue-life calculations depending on the guide and preload condition.

Do Roller and Ball Guides Use the Same Life Calculation?

Not always. Their fatigue-life equations and exponents can differ, so use the manufacturer's specified method.

Need Help Calculating Linear Guide Load?

For a useful application review, record:

total moving mass,
individual payload masses,
center-of-gravity location,
rail spacing,
block spacing,
number of rails,
number of blocks,
mounting orientation,
maximum speed,
acceleration,
deceleration,
external/process forces,
stroke,
duty cycle,
required life,
desired rigidity,
desired accuracy.

For an existing machine, also provide:

rail manufacturer,
full rail part number,
full carriage part number,
rail size,
photographs.

Contact Linear Automation USA with this information when you need help evaluating an industrial linear guide application.