How to Compare Airbag Designs Using Force-Time Graphs

Two airbags can produce very different force-time graphs even when they stop the same crash-test dummy. One might produce a tall, narrow force pulse, while another produces a lower, wider pulse. Which design is better?

You cannot decide from the peak force alone. To compare airbag designs properly, examine three things: the area under each force-time graph, the mean force during the collision, and the peak force.

If both tests bring identical dummies from the same initial velocity to the same final velocity, the total impulse must be the same. The more useful comparison is how that impulse is delivered over time.

01Start with the area under the graph

Imagine a 50 kg crash-test dummy travelling forwards at 8.0 m s⁻¹. An airbag brings it to rest.

Its initial momentum is:

\[ p_i=mv_i \]

\[ p_i=50\times8.0=400\text{ kg m s}^{-1} \]

Its final momentum is zero, so the change in momentum is:

\[ \Delta p=0-400=-400\text{ kg m s}^{-1} \]

The negative sign indicates that the momentum change is backwards.

Therefore, the total restraining force must deliver an impulse of magnitude:

\[ |J|=400\text{ N s} \]

Now predict something: if a second airbag stops the same dummy under identical conditions, should its force-time graph have the same area?

Yes. The dummy has the same mass and undergoes the same change in momentum. Therefore, the total net impulse must also be 400 N s.

The graphs can have completely different shapes, but their signed areas must match.

Comparing two illustrative airbag designs

Suppose two idealised triangular force-time pulses are recorded. Both represent the total horizontal restraining force acting on the dummy.

• Design A produces a peak force of 20,000 N over 0.040 s.
• Design B produces a peak force of 10,000 N over 0.080 s.

The areas are:

For Design A:

\[ J_A=\frac12(20\,000)(0.040) \]

\[ J_A=400\text{ N s} \]

For Design B:

\[ J_B=\frac12(10\,000)(0.080) \]

\[ J_B=400\text{ N s} \]

Both designs deliver the same impulse.

However, their mean and peak forces are different.

Design B has half the peak force and half the mean force, despite producing exactly the same impulse.

This is the central idea behind comparing airbag designs using force-time graphs: a longer interaction can deliver the required momentum change with a smaller average force.

The lower peak is also useful evidence, but you must read it directly from the graph rather than assume that a longer collision automatically produces a lower peak.

02Area, mean force, and peak force are different measurements

Each measurement answers a different question.

1. Area: how much impulse is delivered?

The area under a force-time graph represents impulse.

\[ J=\Delta p=m(v_f-v_i) \]

Here:

• \(J\) is impulse, measured in newton seconds (N s).
• \(m\) is the dummy’s mass in kilograms (kg).
• \(v_i\) is its initial velocity in metres per second (m s⁻¹).
• \(v_f\) is its final velocity in metres per second (m s⁻¹).
• \(\Delta p\) is the change in momentum.

For a graph made from straight-line sections, calculate the area by splitting it into familiar shapes:

• Rectangle: \(A=bh\)
• Triangle: \(A=\frac12 bh\)
• Trapezium: \(A=\frac12(a+b)h\)

Remember to convert milliseconds into seconds before calculating impulse.

The area tells you the total impulse, not the maximum force experienced during the collision.

2. Mean force: how large is the average force?

The mean force is the constant force that would produce the same impulse over the same time interval.

\[ F_{\text{mean}}=\frac{|J|}{\Delta t} \]

Here, \(\Delta t\) is the duration of the collision in seconds.

For Design A:

\[ F_{\text{mean}}=\frac{400}{0.040} \]

\[ F_{\text{mean}}=10\,000\text{ N} \]

For Design B:

\[ F_{\text{mean}}=\frac{400}{0.080} \]

\[ F_{\text{mean}}=5\,000\text{ N} \]

The mean force can be pictured as the height of a rectangle with the same area and duration as the actual force-time pulse.

The dummy does not necessarily experience this force at any particular instant. It is an average across the interaction.

3. Peak force: what is the largest instantaneous force?

The peak force is simply the highest force magnitude reached during the collision.

For our two designs:

• Design A: 20,000 N.
• Design B: 10,000 N.

Peak force and mean force are not interchangeable.

For a triangular pulse that begins and ends at zero, the mean force magnitude is half the peak force. However, this relationship does not automatically apply to rectangular or trapezium-shaped pulses.

For other shapes, calculate the area and divide by the duration.

This distinction is particularly important when an exam question gives you a force-time graph rather than a simple triangular pulse.

If you need to revisit why graph area represents impulse, see Force-Time Graphs: Impulse and Momentum Change in HSC Physics.

03How would you fairly test two airbag designs?

There is an experimental problem hiding behind this comparison.

Suppose an engineer wants to determine whether a redesigned airbag reduces the force experienced by an occupant. How can they tell whether the improvement comes from the airbag rather than a difference in the crash conditions?

A controlled crash-test experiment provides a way to investigate this.

Setting up the apparatus

A simplified crash-test arrangement uses a sled-mounted seat.

The seat is attached to a sled that travels along a horizontal track. An instrumented crash-test dummy sits in the seat, secured by a seatbelt.

An airbag module is mounted directly in front of the dummy, such as within a steering-wheel or dashboard assembly.

Sensors are placed in the dummy and restraint system. An accelerometer records the dummy’s acceleration, while suitable force sensors can record loads from individual restraints.

The sensors connect to a data logger, which records measurements throughout the collision.

What should the experiment show?

The experiment investigates whether a different airbag produces a more favourable force-time profile under otherwise identical conditions.

To make the comparison fair, keep the following conditions consistent:

• Dummy mass and dimensions.
• Initial speed of the sled and dummy.
• Seat position and dummy posture.
• Seatbelt adjustment.
• Airbag position and deployment timing.
• Sled deceleration profile.

The prediction is that a restraint system that increases the time taken to stop the dummy can reduce its mean force, provided the required momentum change remains the same.

However, a longer stopping time does not guarantee a lower peak. The entire force-time curve must be recorded.

What does the equipment actually measure?

There is an important distinction between measuring the force from the airbag and measuring the net force on the dummy.

In a simplified horizontal model, Newton’s second law gives:

\[ F_{\text{net}}=ma \]

Here, \(F_{\text{net}}\) is the resultant horizontal force, \(m\) is the dummy’s mass, and \(a\) is its horizontal acceleration.

An accelerometer can therefore be used to estimate the net horizontal force.

A force sensor measuring the airbag’s contact force, however, records only that particular interaction. The seatbelt may also exert a substantial force.

These measurements are not interchangeable.

For a complete comparison, either examine the resultant force on the dummy or account for all relevant restraint forces.

Why measuring only the maximum force is insufficient

Imagine an initial test that records only the dummy’s maximum acceleration.

This tells the investigator something about the peak resultant force, but it does not reveal how long the force acts.

A short, sharp force pulse and a broad, sustained pulse could produce very different impulses despite having the same peak.

The measurement method needs to be improved.

Recording the entire force-time trace allows the investigator to calculate:

1. The total impulse from the area.
2. The mean force from the impulse and duration.
3. The peak force from the highest point on the graph.

Initial and final velocity measurements provide an additional check using the change in momentum.

The result is a much more useful comparison than a single maximum-force reading.

All numerical data in this article are illustrative teaching values, not measurements from a particular crash test or specifications for commercial airbags.

04The lower-peak misconception

There is a tempting mistake when comparing two airbag designs.

A student sees that one design has a smaller peak force and immediately concludes that it must be better.

But what if its force-time graph has a smaller area as well?

Consider the same 50 kg dummy travelling at 8.0 m s⁻¹.

It requires an impulse of 400 N s to come to rest.

Now consider another triangular force pulse:

• Duration: 0.080 s.
• Peak force: 7,500 N.

Its impulse is:

\[ J=\frac12(7\,500)(0.080) \]

\[ J=300\text{ N s} \]

This peak is lower than either of the previous designs. However, the impulse is also smaller.

What happens to the dummy?

Its initial momentum is:

\[ p_i=50(8.0)=400\text{ kg m s}^{-1} \]

Taking forward as positive, the backward impulse is \(-300\text{ N s}\).

Therefore:

\[ p_f=p_i+J \]

\[ p_f=400-300 \]

\[ p_f=100\text{ kg m s}^{-1} \]

Its final velocity is:

\[ v_f=\frac{p_f}{m} \]

\[ v_f=\frac{100}{50} \]

\[ v_f=2.0\text{ m s}^{-1} \]

The dummy is still moving forwards at 2.0 m s⁻¹.

The smaller peak does not demonstrate a better stopping process. The dummy has not undergone the required momentum change during the recorded interaction.

There is one important qualification: this conclusion assumes the graph represents the total net horizontal force throughout the entire interaction.

If the graph represents only the airbag’s force, the seatbelt could supply the remaining impulse.

This is why identifying exactly which force is being graphed is essential. The distinction between an individual interaction force and the net force is also explored in Can a Collision Have Zero Net Impulse but a Large Force?.

05A reliable method for comparing airbag designs

When you encounter an airbag comparison question in an HSC Physics exam, follow this sequence.

Step 1: Identify the force being measured.

Check whether the graph represents the airbag’s contact force, the seatbelt force, or the total net force on the dummy.

Step 2: Check whether the tests are comparable.

Look for the same dummy mass, initial velocity, final velocity, and crash conditions.

If the momentum changes are different, the required impulses will also be different.

Step 3: Calculate the area under each graph.

Split the graph into triangles, rectangles, and trapeziums.

Convert all times into seconds and calculate the signed impulse.

Step 4: Compare the mean forces.

Use:

\[ F_{\text{mean}}=\frac{|J|}{\Delta t} \]

A longer stopping interval can reduce the mean force when the impulse is unchanged.

Step 5: Read the peak forces.

Identify the largest force magnitude reached by each design.

Do not assume that a lower mean force automatically means a lower peak force.

Step 6: Make a qualified conclusion.

Explain whether the designs deliver the same impulse, and compare their mean and peak forces.

If the data only describe one restraint component, acknowledge that other forces may contribute to the dummy’s total momentum change.

A strong HSC response explains both the numerical comparison and its physical meaning.

06Worked examples

07Questions and solutions

Question 1

A 60 kg crash-test dummy initially travels forwards at 6.0 m s⁻¹. It is brought to rest by a triangular total restraining-force pulse lasting 0.050 s.

Assuming the force acts backwards throughout the collision, calculate:

a) The magnitude of the impulse.

b) The mean force magnitude.

c) The peak force magnitude.

Solution 1

The impulse magnitude is 360 N s, the mean force is 7.2 kN, and the peak force is 14.4 kN.

a) Impulse

Use the change in momentum:

\[ |J|=m|v_f-v_i| \]

Substitute:

\[ |J|=60|0-6.0| \]

\[ |J|=360\text{ N s} \]

b) Mean force

Convert the duration:

\[ 0.050\text{ s}=0.050\text{ s} \]

Then:

\[ F_{\text{mean}}=\frac{|J|}{\Delta t} \]

\[ F_{\text{mean}}=\frac{360}{0.050} \]

\[ F_{\text{mean}}=7\,200\text{ N} \]

c) Peak force

Because the graph is triangular:

\[ J=\frac12 F_{\text{peak}}\Delta t \]

Rearrange:

\[ F_{\text{peak}}=\frac{2J}{\Delta t} \]

Substitute:

\[ F_{\text{peak}}=\frac{2(360)}{0.050} \]

\[ F_{\text{peak}}=14\,400\text{ N} \]

The peak force is twice the mean force because the pulse is triangular.

Question 2

A 54 kg crash-test dummy travels forwards at 10 m s⁻¹ before being brought to rest.

Two airbag designs produce the following total horizontal restraining-force profiles.

Design A

• Rises from zero to 18 kN over 10 ms.
• Remains at 18 kN for 20 ms.
• Falls to zero over 10 ms.

Design B

• Rises from zero to 12 kN over 20 ms.
• Remains at 12 kN for 25 ms.
• Falls to zero over 20 ms.

Calculate the impulse, mean force, and peak force for each design.

Explain why the two impulse values should agree, and identify which design has the lower mean and peak forces.

Solution 2

Design B has the lower mean force and peak force. Both designs deliver an impulse of 540 N s, as required to bring the dummy to rest.

Step 1: Calculate the required impulse.

\[ |J|=m|v_f-v_i| \]

\[ |J|=54|0-10| \]

\[ |J|=540\text{ N s} \]

Step 2: Calculate the area for Design A.

First triangle:

\[ J_1=\frac12(0.010)(18\,000) \]

\[ J_1=90\text{ N s} \]

Rectangle:

\[ J_2=(0.020)(18\,000) \]

\[ J_2=360\text{ N s} \]

Second triangle:

\[ J_3=\frac12(0.010)(18\,000) \]

\[ J_3=90\text{ N s} \]

Total:

\[ J_A=90+360+90 \]

\[ J_A=540\text{ N s} \]

Step 3: Calculate the mean and peak forces for Design A.

The duration is:

\[ \Delta t_A=0.010+0.020+0.010 \]

\[ \Delta t_A=0.040\text{ s} \]

Mean force:

\[ F_{\text{mean,A}}=\frac{540}{0.040} \]

\[ F_{\text{mean,A}}=13\,500\text{ N} \]

Peak force:

\[ F_{\text{peak,A}}=18\,000\text{ N} \]

Step 4: Calculate the area for Design B.

First triangle:

\[ J_1=\frac12(0.020)(12\,000) \]

\[ J_1=120\text{ N s} \]

Rectangle:

\[ J_2=(0.025)(12\,000) \]

\[ J_2=300\text{ N s} \]

Second triangle:

\[ J_3=\frac12(0.020)(12\,000) \]

\[ J_3=120\text{ N s} \]

Total:

\[ J_B=120+300+120 \]

\[ J_B=540\text{ N s} \]

Step 5: Calculate the mean and peak forces for Design B.

The duration is:

\[ \Delta t_B=0.020+0.025+0.020 \]

\[ \Delta t_B=0.065\text{ s} \]

Mean force:

\[ F_{\text{mean,B}}=\frac{540}{0.065} \]

\[ F_{\text{mean,B}}\approx8\,308\text{ N} \]

Peak force:

\[ F_{\text{peak,B}}=12\,000\text{ N} \]

Step 6: Compare the designs.

The impulse values agree because the dummy undergoes the same momentum change in both tests.

Design B spreads that impulse over a longer period, producing a lower mean and peak force.

Question 3

A 54 kg crash-test dummy travels forwards at 10 m s⁻¹ and is eventually brought to rest.

During the collision, an airbag exerts a backward force that increases uniformly from zero to 12,000 N before decreasing uniformly to zero. The interaction lasts 0.050 s.

A separate seatbelt also exerts a backward force on the dummy.

Assuming other horizontal impulses are negligible, calculate:

a) The impulse delivered by the airbag.

b) The impulse that must be delivered by the seatbelt.

c) Explain why the area under the airbag force-time graph does not equal the dummy’s total change in momentum.

Solution 3

The airbag supplies a backward impulse of 300 N s, while the seatbelt must supply a further backward impulse of 240 N s.

a) Airbag impulse

The force-time graph is triangular.

\[ J_{\text{airbag}}=\frac12 F_{\text{peak}}\Delta t \]

Substitute:

\[ J_{\text{airbag}} =\frac12(12\,000)(0.050) \]

\[ J_{\text{airbag}}=300\text{ N s} \]

The signed impulse is \(-300\text{ N s}\), because the force acts backwards.

b) Seatbelt impulse

First calculate the total momentum change required to stop the dummy.

\[ \Delta p=m(v_f-v_i) \]

\[ \Delta p=54(0-10) \]

\[ \Delta p=-540\text{ kg m s}^{-1} \]

Therefore, the total backward impulse must be \(-540\text{ N s}\).

The two restraint impulses add:

\[ J_{\text{total}}=J_{\text{airbag}}+J_{\text{belt}} \]

\[ -540=-300+J_{\text{belt}} \]

\[ J_{\text{belt}}=-240\text{ N s} \]

The seatbelt supplies a backward impulse of magnitude 240 N s.

c) Why are the areas different?

The airbag graph represents only the force exerted by the airbag.

The dummy also experiences a force from the seatbelt. Therefore, the area under the airbag graph represents only the airbag’s contribution to the total impulse.

The total momentum change depends on the net impulse from all relevant forces.

This is why identifying the force represented by a graph is just as important as calculating its area.

Question 4

A 50 kg crash-test dummy initially travels forwards at 8.0 m s⁻¹.

Two candidate restraint systems produce triangular total horizontal restraining-force pulses.

Design C

• Duration: 0.080 s.
• Peak force: 7,500 N.

Design D

• Duration: 0.080 s.
• Peak force: 10,000 N.

A student claims that Design C must be better because it produces a smaller peak force.

a) Calculate the impulse delivered by each design.

b) Calculate the mean force for each design.

c) Calculate the dummy’s final velocity after experiencing Design C.

d) Explain why the student’s conclusion is incorrect.

Assume the graphs cover the complete horizontal restraint interaction.

Solution 4

Design C has a smaller peak and mean force, but it does not deliver enough impulse to stop the dummy. Its final velocity is 2.0 m s⁻¹ forwards. Therefore, the student’s conclusion is incorrect.

a) Calculate the impulses.

For Design C:

\[ J_C=\frac12 F_{\text{peak}}\Delta t \]

\[ J_C=\frac12(7\,500)(0.080) \]

\[ J_C=300\text{ N s} \]

For Design D:

\[ J_D=\frac12(10\,000)(0.080) \]

\[ J_D=400\text{ N s} \]

b) Calculate the mean forces.

For Design C:

\[ F_{\text{mean,C}}=\frac{300}{0.080} \]

\[ F_{\text{mean,C}}=3\,750\text{ N} \]

For Design D:

\[ F_{\text{mean,D}}=\frac{400}{0.080} \]

\[ F_{\text{mean,D}}=5\,000\text{ N} \]

Design C has the smaller mean and peak forces.

However, that does not establish that it has completed the stopping process.

c) Calculate the final velocity after Design C.

The initial momentum is:

\[ p_i=mv_i \]

\[ p_i=50(8.0) \]

\[ p_i=400\text{ kg m s}^{-1} \]

The impulse acts backwards:

\[ J_C=-300\text{ N s} \]

Therefore:

\[ p_f=p_i+J_C \]

\[ p_f=400-300 \]

\[ p_f=100\text{ kg m s}^{-1} \]

The final velocity is:

\[ v_f=\frac{p_f}{m} \]

\[ v_f=\frac{100}{50} \]

\[ v_f=2.0\text{ m s}^{-1} \]

The dummy is still moving forwards.

Design D, on the other hand, supplies the full 400 N s needed to bring the dummy to rest.

d) Explain the misconception.

The student has compared peak forces without checking whether the designs produce the same momentum change.

Design C has a smaller peak because its total area is smaller. It has not delivered enough impulse to stop the dummy.

A valid comparison must account for the required change in momentum before interpreting a smaller peak as an improvement.

Question 5

This final problem combines impulse, mean force, peak force, and the interaction between different restraints.

A 60 kg crash-test dummy initially travels forwards at 10 m s⁻¹. Two restraint systems are tested under otherwise identical conditions. Both bring the dummy to rest.

In each test, the total horizontal restraining force is the sum of the airbag force and seatbelt force. Both forces act backwards.

The sensors record the following idealised rectangular force pulses.

Trial A

• Airbag force: 12,000 N from 0 to 25 ms.
• Seatbelt force: 6,000 N from 25 to 75 ms.

Trial B

• Airbag force: 8,000 N from 0 to 37.5 ms.
• Seatbelt force: 6,000 N from 0 to 50 ms.

Assume that no other horizontal forces contribute significantly.

A student argues that Trial B must be better because its airbag peak force is lower.

a) Calculate the impulse delivered by the airbag and seatbelt separately in each trial.

b) Determine the total peak force in each trial.

c) Calculate the mean total restraining force in each trial.

d) Explain why comparing the airbag peak forces alone gives the wrong conclusion.

e) Which trial has the lower total peak and mean force? Explain what can and cannot be concluded from these results.

Solution 5

Trial A has a total peak force of 12 kN and a mean force of 8 kN. Trial B has a total peak force of 14 kN and a mean force of 12 kN.

Although Trial B has the smaller airbag peak, the overlap between its airbag and seatbelt forces produces a larger total peak.

Step 1: Calculate the required total impulse.

The dummy starts at 10 m s⁻¹ and finishes at rest.

\[ |J|=m|v_f-v_i| \]

\[ |J|=60|0-10| \]

\[ |J|=600\text{ N s} \]

Both trials must deliver a total backward impulse of 600 N s.

Step 2: Calculate the individual impulses in Trial A.

Airbag:

\[ J_{\text{airbag,A}}=(12\,000)(0.025) \]

\[ J_{\text{airbag,A}}=300\text{ N s} \]

Seatbelt:

\[ J_{\text{belt,A}}=(6\,000)(0.050) \]

\[ J_{\text{belt,A}}=300\text{ N s} \]

Total:

\[ J_A=300+300 \]

\[ J_A=600\text{ N s} \]

Step 3: Calculate the total peak force in Trial A.

The airbag acts from 0 to 25 ms.

The seatbelt acts from 25 to 75 ms.

The forces do not overlap.

Therefore, the total force is:

• 12,000 N from 0 to 25 ms.
• 6,000 N from 25 to 75 ms.

The highest total force is:

\[ F_{\text{peak,A}}=12\,000\text{ N} \]

Step 4: Calculate the mean force in Trial A.

The complete interaction lasts 75 ms.

\[ \Delta t_A=0.075\text{ s} \]

\[ F_{\text{mean,A}}=\frac{600}{0.075} \]

\[ F_{\text{mean,A}}=8\,000\text{ N} \]

Step 5: Calculate the individual impulses in Trial B.

Airbag:

\[ J_{\text{airbag,B}}=(8\,000)(0.0375) \]

\[ J_{\text{airbag,B}}=300\text{ N s} \]

Seatbelt:

\[ J_{\text{belt,B}}=(6\,000)(0.050) \]

\[ J_{\text{belt,B}}=300\text{ N s} \]

Total:

\[ J_B=300+300 \]

\[ J_B=600\text{ N s} \]

Both trials therefore deliver the same total impulse.

Step 6: Calculate the total peak force in Trial B.

This is where the comparison becomes interesting.

The airbag and seatbelt forces act simultaneously from 0 to 37.5 ms.

Because both forces act backwards, their magnitudes add:

\[ F_{\text{total}}=8\,000+6\,000 \]

\[ F_{\text{total}}=14\,000\text{ N} \]

From 37.5 to 50 ms, only the seatbelt force remains, at 6,000 N.

Therefore:

\[ F_{\text{peak,B}}=14\,000\text{ N} \]

The smaller airbag peak has actually produced a larger total peak force.

Step 7: Calculate the mean force in Trial B.

The complete interaction lasts 50 ms.

\[ \Delta t_B=0.050\text{ s} \]

\[ F_{\text{mean,B}}=\frac{600}{0.050} \]

\[ F_{\text{mean,B}}=12\,000\text{ N} \]

We can also check the total area by considering the combined force profile.

During the first 37.5 ms, both forces act:

\[ J_1=(14\,000)(0.0375) \]

\[ J_1=525\text{ N s} \]

During the remaining 12.5 ms, only the belt acts:

\[ J_2=(6\,000)(0.0125) \]

\[ J_2=75\text{ N s} \]

Therefore:

\[ J_B=525+75=600\text{ N s} \]

The area agrees with the required momentum change.

Step 8: Compare the final results.

Trial A has the lower total peak and mean forces.

The important detail is that the total force depends on when the individual restraint forces act.

Adding the individual peak forces without considering their timing would give the wrong answer. Equally, comparing only the airbag peaks ignores the contribution from the seatbelt.

This is a more complete comparison because it considers the combined force experienced by the dummy.

However, it still does not establish which system would cause fewer injuries. The distribution of force across the body, contact area, and other injury-related measurements would also need to be considered.

08What to examine next

A force-time graph tells you how the total restraining force changes throughout a collision. It also lets you compare impulse, mean force, and peak force without confusing these three measurements.

But force is not the only factor affecting the load on an occupant.

An airbag also spreads the contact force over an area. This introduces pressure:

\[ P=\frac{F}{A} \]

where \(P\) is pressure in pascals (Pa), \(F\) is the applied force in newtons (N), and \(A\) is the contact area in square metres (m²).

The next useful step is to consider how force and contact area work together. A force-time comparison tells you how the restraint changes the occupant’s momentum; pressure helps explain how that force is distributed across the body.