Strapdown Inertial Navigation Examples: Specific Force and Accelerometer Physics

This companion to INS03 turns the specific-force equation into a working calculation tool. The central relationship is

(1)

or equivalently

(2)

The exercises deliberately mix ordinary mechanics, coordinate transformations, static leveling, and navigation implementation checks. The purpose is to make the reader comfortable answering two different questions:

1. What will an ideal accelerometer report in a given physical situation?
2. Given an accelerometer measurement, what additional information is required to reconstruct vehicle acceleration?

Unless otherwise stated, use

(3)

When standard-gravity units are requested, use

(4)

The notation follows INS03 and standard inertial-navigation references [1, 2, 3, 4]. Body axes use the aerospace forward-right-down convention when explicitly stated, and the local navigation frame is North-East-Down (NED).

1 Exercises

Exercise 1: A supported proof mass on a table

A small ideal accelerometer has a proof mass

(5)

It rests motionless on a horizontal table. Use a local Cartesian frame with +z upward.

1. Write the gravitational acceleration vector g.
2. Find the inertial acceleration a of the accelerometer.
3. Find the specific force f.
4. Find the total non-gravitational force acting on the proof mass.
5. Express the accelerometer output magnitude in units of g₀.
6. Explain why the answer is nonzero although the accelerometer is motionless.

Exercise 2: Elevator readings and apparent weight

A 75.0 kg passenger stands on a scale in an elevator. An ideal accelerometer is rigidly attached to the elevator. Again take +z upward.

For each of the following cases, find the accelerometer output f_z and the scale Normal force N:

1. constant velocity;
2. upward acceleration a_z = +2.50 m∕s²;
3. downward acceleration a_z = −3.00 m∕s²;
4. ideal cable-break free fall.

Explain why the accelerometer and the scale tell essentially the same mechanical story.

Exercise 3: Engine cutoff during an upward flight

A vehicle is moving vertically upward at 500 m∕s near the Earth’s surface. At one instant its engine is shut off and aerodynamic drag is neglected.

1. Immediately after engine cutoff, what is its coordinate acceleration?
2. What does its ideal accelerometer read?
3. Does the fact that the vehicle is moving upward at 500 m∕s change the accelerometer result?
4. Later, suppose the engine produces a net vehicle acceleration of +15.0 m∕s² upward. What specific force must then be measured?

Use the result to explain why zero accelerometer output does not mean zero coordinate acceleration.

Exercise 4: Level car accelerating horizontally

A CAR travels on a level road and accelerates eastward at

(6)

Use a local frame with +x east and +z upward.

1. Write a and g.
2. Compute the specific-force vector f.
3. Find |f|.
4. Find the angle of f from the upward vertical.
5. Identify the physical non-gravitational interactions producing the horizontal and vertical components.

Exercise 5: Body-frame accelerometer data to NED acceleration

A level vehicle uses forward-right-down body axes. Its heading is

(7)

measured east of north. Its ideal accelerometers report

(8)

Assume the vehicle is level and neglect Earth rotation for this exercise.

1. Construct the body-to-NED DCM C_bⁿ.
2. Compute fⁿ = C _bⁿf^b.
3. Add the NED gravity vector

   to recover aⁿ.

4. Interpret the north, east, and down components of the result.

Exercise 6: Static leveling from accelerometer components

A stationary IMU uses forward-right-down body axes. Its true roll and pitch are

(9)

Yaw is arbitrary. Assume the local navigation frame is NED and

(10)

For the usual yaw-pitch-roll attitude convention, show that a static accelerometer satisfies

(11)

(12)

(13)

Then:

1. compute f_x,f_y,f_z numerically;
2. recover pitch from

3. recover roll from

4. explain why yaw cannot be obtained from the accelerometer vector alone.

Exercise 7: Small attitude error produces false horizontal acceleration

A vehicle is actually stationary and level. Its accelerometers are perfect, but the navigation computer’s attitude estimate has a constant pitch error of

(14)

Assume the only consequence is that the vertical specific-force vector is rotated incorrectly into the navigation frame.

1. Find the exact magnitude of the false horizontal acceleration using g sin δ𝜃.
2. Compare it with the small-angle approximation gδ𝜃 when δ𝜃 is in radians.
3. If this false acceleration remains constant for 60 s, estimate the resulting horizontal velocity error.
4. Estimate the horizontal position error after 60 s.
5. Explain why attitude accuracy is inseparable from accelerometer accuracy in a strapdown INS.

Exercise 8: Accelerometer on a horizontal rotating arm

An accelerometer is mounted at radius

(15)

on a horizontal arm rotating at constant angular speed

(16)

At the instant of interest define +x radially outward and +z upward. Neglect Earth rotation but retain gravity.

1. Find the coordinate acceleration a.
2. Compute the specific force f = a − g.
3. Find its magnitude in m∕s² and in units of g ₀.
4. Find the angle of the measured specific-force vector away from the upward vertical toward the rotation axis.
5. Explain which physical forces produce the two measured components.

Exercise 9: A pitched vehicle – complete specific-force reconstruction

A vehicle uses forward-right-down body axes and NED navigation coordinates. At one instant it has zero roll and yaw and a nose-up pitch angle

(17)

The body-frame accelerometer vector is

(18)

Use

(19)

Neglect Earth rotation.

1. Compute fⁿ.
2. Add gⁿ = [0 0 9.81]^T  m∕s².
3. Find the north and vertical coordinate accelerations.
4. State whether the vehicle is accelerating upward or downward.

Exercise 10: Navigation-code sanity checks

A stationary, level IMU in NED coordinates ideally has

(20)

A programmer makes two different mistakes.

1. Code A treats fⁿ directly as coordinate acceleration. What down-velocity and vertical-position error does it accumulate after 30 s from rest?
2. Code B computes aⁿ = fⁿ − gⁿ instead of adding gravity. What down-velocity and vertical-position error does it accumulate after 30 s?
3. State the correct stationary result.
4. Propose at least four unit tests that should be placed around a strapdown specific-force implementation.

2 Worked solutions

Solution 1: A supported proof mass on a table

With +z upward,

(21)

The accelerometer is stationary, so

(22)

Specific force is

f	= a − g	(23)
	= 0 − (−9.81ez),	(24)

hence

(25)

The proof mass is

(26)

Because f = F_ng∕m,

Fng	= mf	(27)
	= (0.0200)(9.81)ez,	(28)

so

(29)

In units of standard gravity,

(30)

The reading is nonzero because the proof mass is not allowed to follow its natural gravitational free-fall trajectory. The sensor structure supplies an upward non-gravitational force that keeps the mass fixed relative to the case. The accelerometer senses that support interaction.

Solution 2: Elevator readings and apparent weight

For vertical motion with +z upward,

(31)

The passenger’s scale normal force satisfies

(32)

so

(33)

Thus the accelerometer’s specific force and the passenger’s scale force per unit mass are the same mechanical quantity in this idealized one-dimensional problem.

For constant velocity,

(34)

so

(35)

and

(36)

For upward acceleration a_z = +2.50 m∕s²,

(37)

and

(38)

For downward acceleration a_z = −3.00 m∕s²,

(39)

and

(40)

In ideal cable-break free fall,

(41)

so

(42)

This is why a falling elevator occupant feels weightless: both the person and the elevator are following nearly the same gravitational trajectory, so the support interaction vanishes.

Solution 3: Engine cutoff during an upward flight

Immediately after engine cutoff, gravity remains. Neglecting drag,

(43)

Because

(44)

specific force is

(45)

The upward velocity of 500 m∕s does not appear in the ideal specific-force relation. Velocity affects the trajectory, but with the stated assumptions it does not create a non-gravitational force. At the instant after shutdown, the vehicle is ballistic and its ideal accelerometer reads zero.

When the engine later produces a net coordinate acceleration of +15.0 m∕s² upward,

fz	= az − (−g)	(46)
	= 15.0 + 9.81,	(47)

so

(48)

This example separates velocity, coordinate acceleration, and accelerometer output. They are three different quantities.

Solution 4: Level car accelerating horizontally

The coordinate acceleration is

(49)

while gravity is

(50)

Therefore

f	= a − g	(51)
	= m∕s2.	(52)

Thus

(53)

Its magnitude is

|f|	=	(54)
	≈ 10.2585 m∕s2 .	(55)

The angle away from the upward vertical is

(56)

The vertical component comes from the road supporting the vehicle against gravity. The horizontal component comes from tire-road traction transmitted through the vehicle structure. Both are non-gravitational interactions and therefore both appear in the accelerometer measurement.

Solution 5: Body-frame accelerometer data to NED acceleration

For a level vehicle with heading ψ, the body x axis points in the horizontal direction

(57)

when resolved in NED coordinates. The right body axis resolves as

(58)

Therefore

(59)

At ψ = 30^∘,

(60)

Transform the accelerometer vector:

fn	= C bnfb	(61)
	= .	(62)

Hence

(63)

Now restore gravity:

an	= fn + gn	(64)
	= m∕s2.	(65)

Therefore

(66)

The horizontal acceleration magnitude is 2.00 m∕s², exactly equal to the forward specific-force component, and it points 30^∘ east of north. The −9.81 m∕s² down component of specific force cancels the +9.81 m∕s² gravity vector because the vehicle is level and has no vertical acceleration.

This is the basic strapdown sequence:

(67)

Solution 6: Static leveling from accelerometer components

For the usual yaw-pitch-roll convention, the body-to-NED DCM has third row

(68)

Because

(69)

and

(70)

the body components are minus g times the third row of C_bⁿ:

(71)

(72)

(73)

With

(74)

we obtain

fx	= 9.81 sin(−5∘)	(75)
	≈−0.8550 m∕s2 ,	(76)

fy	= −9.81 sin(10∘) cos(−5∘)	(77)
	≈−1.6970 m∕s2 ,	(78)

and

fz	= −9.81 cos(10∘) cos(−5∘)	(79)
	≈−9.6242 m∕s2 .	(80)

The magnitude remains

(81)

as expected for an ideal stationary accelerometer in the assumed uniform field.

Recover pitch:

𝜃	= sin −1	(82)
	≈−5.00∘ .	(83)

Recover roll:

ϕ	= atan2(−fy,−fz)	(84)
	= atan2(1.6970, 9.6242)	(85)
	≈ 10.00∘ .	(86)

Yaw is not observable because rotating the vehicle about the local vertical does not change the direction of gravity relative to that vertical. The accelerometer provides a vertical reference, not a north reference.

Solution 7: Small attitude error produces false horizontal acceleration

Convert the attitude error to radians:

(87)

The exact horizontal projection of the vertical specific-force vector is

δaH	= g sin δ𝜃	(88)
	= 9.81 sin(0.50∘)	(89)
	≈ 0.0856073 m∕s2 .	(90)

The small-angle approximation gives

gδ𝜃	= (9.81)(0.00872665)	(91)
	≈ 0.0856084 m∕s2 .	(92)

The difference is only about 1.1 × 10⁻⁶ m∕s² here, so the linear approximation is excellent.

If the false acceleration is constant for t = 60 s,

(93)

The corresponding position error from rest is

δrH	= δaHt2	(94)
	≈(0.0856073)(60)2	(95)
	≈ 154.1 m .	(96)

This result is fundamental to strapdown navigation. The accelerometer itself can be perfect, but if attitude is wrong, the navigation computer resolves the large vertical support vector into the wrong directions. Attitude errors therefore become acceleration errors before the first velocity integration even occurs.

Solution 8: Accelerometer on a horizontal rotating arm

The centripetal coordinate acceleration is directed inward. Since +x is radially outward,

(97)

Numerically,

(98)

so

(99)

Gravity is

(100)

Therefore

f	= a − g	(101)
	= −4.50ex + 9.81ez.	(102)

Thus

(103)

Its magnitude is

|f|	=	(104)
	≈ 10.7929 m∕s2 .	(105)

In standard-gravity units,

(106)

The angle from the upward vertical toward the rotation axis is

(107)

The upward component is generated by the structural support against gravity. The inward component is generated by the arm supplying the centripetal force needed to keep the accelerometer on its circular path. Both are non-gravitational, so both are sensed.

Solution 9: A pitched vehicle – complete specific-force reconstruction

With 𝜃 = 20^∘,

(108)

Therefore

fn	= C bnfb	(109)
	= .	(110)

The north component is

fN	= (0.939693)(12.0) + (0.342020)(−9.0)	(111)
	≈ 8.1981 m∕s2 .	(112)

The east component is zero. The down component is

fD	= (−0.342020)(12.0) + (0.939693)(−9.0)	(113)
	≈−12.5615 m∕s2 .	(114)

Hence

(115)

Restore gravity:

an	= fn + gn	(116)
	= +	(117)
	= m∕s2.	(118)

Therefore

(119)

and

(120)

Because NED takes positive D downward, negative down acceleration means the vehicle is accelerating upward at approximately

(121)

This example shows why one must transform the full specific-force vector before adding gravity. Gravity cannot simply be added to the body z channel unless body and navigation vertical axes are actually aligned.

Solution 10: Navigation-code sanity checks

The correct stationary relation is

(122)

Code A incorrectly uses

(123)

Starting from rest, after 30 s,

(124)

The down-position error is

ΔDA	= (−9.81)(30)2	(125)
	= −4414.5 m .	(126)

A negative down displacement corresponds to an erroneous altitude increase of 4414.5 m.

Code B computes

aBn	= fn − gn	(127)
	= m∕s2.	(128)

Thus

(129)

and

ΔDB	= (−19.62)(30)2	(130)
	= −8829 m .	(131)

The correct stationary result is

(132)

apart from whatever Earth-rotation and gravity-model details are included in the chosen full mechanization.

Useful unit tests include:

1. Stationary level NED test:

   must produce zero translational acceleration.

2. Ideal free-fall test: f = 0 must reconstruct a = g in an inertial-frame calculation.
3. Horizontal acceleration test: a level vehicle with f_N = 2 m∕s² and f _D = −g must reconstruct a_N = 2 m∕s² and a _D = 0.
4. Attitude transformation test: rotate a known body specific-force vector by a known DCM and compare the result with a hand-computed navigation-frame vector.
5. Gravity sign test: in NED, gravity is positive down while the static accelerometer specific force is negative down.
6. norm preservation test: a pure rotation must satisfy

7. Round-trip frame test:

   to numerical precision.

These are inexpensive tests and catch several of the most destructive sign, frame, and interpretation errors before they enter the velocity and position integrations.

3 What these exercises establish

The worked examples reinforce the physical and computational ideas that will recur throughout the strapdown series:

1. An accelerometer measures non-gravitational force per unit mass, not coordinate acceleration directly.
2. A supported stationary accelerometer reads approximately 1g, whereas an ideal freely falling accelerometer reads zero.
3. Velocity by itself does not determine accelerometer output.
4. Horizontal traction and vertical support forces combine vectorially in the accelerometer measurement.
5. Body-frame accelerometer data must be transformed by attitude before gravity can be restored in the navigation frame.
6. Static gravity provides roll and pitch information but not yaw.
7. A small attitude error can project gravity-scale specific force into a horizontal channel and create rapidly growing navigation error.
8. Circular motion produces a horizontal specific-force component. A non-gravitational centripetal force is required to hold the sensor on the circular path.
9. Gravity must be added using the correct frame and sign convention. Treating specific force as coordinate acceleration produces enormous integrated errors.

These results prepare the reader for INS04, which asks the rotational analogue of INS03: what does a gyroscope actually measure?

References