A Dynamic Balance Primer

by Jerry Holway©2003

v.4.1.2003

Table of ContentsOverview 2

Definitions 2

Why is dynamic balance important? 3

Dynamic balance fundamentals 3

The physics and the mathathematics 3 – 11A simple 4 mass object 3A simple 2 mass object, coplanar 4 – 5A simple 2 mass object, non-coplanar 5 – 6A 3 mass object 7 – 9A 5 mass object 10More masses and practical limitations 10 – 11

Why is the math important? 11

A quick and easy method to achieve dynamic balance 12What happens if....? 13

A non-mathematical conceptual model 13What to do if....? 14

Using a dynamic balance program 15

Some things not to do and why 15

A static test for dynamic balance? 16

Trimming for headroom 16

Tilting the monitor 16

Conclusion 17

1

Hey! Skip the math!just read page 12!!!

Overview

In early 1988, GarrettBrown asked Arnold DiGuilioto write an article for thesummer 1988 issue of TheSteadicam Letter on the realmathematics that describe aSteadicam’s behavior.

For the past 15 years,Garrett and I have taughtdynamic balance to hundreds ofoperators, and we have radi-cally improved our practicalmethods and explanations.

Arnold’s mathematicswere complete and accurate.However, Arnold’s wonderfullycompact explanation left manynon-mathematically inclinedfolks scratching their heads.

Many operators have nevereven seen this article, and a lotof confusion and misinforma-tion still exist about dynamicbalance today.

Therefore, it seems to be agood time to offer everyone inthe Steadicam community asolid understanding of dynamicbalance, including a full re-working of the basic physicsand mathematics.

What is dynamic balance?How does one get a sled intodynamic balance? How impor-tant is dynamic balance? Whatare the real physics and math-ematics that explain how aSteadicam behaves, bothstatically and dynamically?

This is an attempt to answerthese questions in a reasonablyfull and complete manner.

First, we need some defini-tions to ensure we are all oncommon ground. Second, I willdiscuss – in some detail – thephysics and mathematics ofdynamic balance.

In the last section, I willdescribe a foolproof and easyway to get any Steadicam intogood dynamic balance and anon-mathematical conceptualmodel of dynamic balance.

Although the math sectionis admittedly a chore to wadethrough, I believe all seriouspractitioners of “the art ofSteadicam” will greatly benefitfrom a thorough understandingof the subject.

Definitions

A Steadicam is in staticbalance when, at rest, it hangswith the central post vertical,regardless of the position of thegimbal on the central post.

An exception: If the sled isperfectly balanced top tobottom, it won’t hang verticallyany more than it will hang atany other angle.

A Steadicam is in dynamicequilibrium if, when rotatedabout the central post, it pansconsistently on that axis.

A Steadicam is in dynamicbalance when it is in dynamicequilibrium and also in staticbalance. Both conditions mustbe met for dynamic balance.

Dynamic balance does notdescribe how fast or slow aSteadicam will pan when aforce is applied. That is aseparate – but related andimportant – subject of inertia.

A Steadicam can often be instatic balance – i.e., it will hangperfectly upright – and not bein dynamic equilibrium, andtherefore it will not be indynamic balance.

In this condition, the rigwill behave very oddly when itis panned. Prior to 1988, this ishow Steadicams were routinelybalanced (and designed!)because, in part, the Steadicamwasn’t considered a quicklyrotating object. Today, opera-tors routinely do whip panswith rotational speeds in theorder of 100 to 150 rpm’s, anddynamic balance is critical forthis type of work.

A minor note: It is possiblefor a Steadicam to be in dy-namic equilibrium and not be instatic balance. While thiscondition is possible to achieveon planet earth, it is useful onlyin outer space when theSteadicam is weightless.

2

Let’s begin our discussionwith two objects that are notSteadicams. This approachmakes it easier to understandthe concepts of static anddynamic balance.

Later we will examine asimple, three mass Steadicam,and we will use the math to putit into dynamic balance.

The physics and the math

The figures below, 1a, 1b,and 1c, represent an object withfour weights on a commonplane at the end of an axle,suspended by a gimbal. Theseweights are “coplanar.”

Why is dynamic balanceimportant?

As an operator, one wantsto be able to frame shots withprecision. Properly balancingthe Steadicam increases theprecision of one's operating.

If a Steadicam is out ofdynamic balance, the operatorconstantly must make adjust-ments to keep the Steadicamlevel as it is panned. Theseadjustments reduce the preci-sion of operating and can affectthe quality or feel of the shot.The more the Steadicam is outof dynamic balance, the greaterthe corrective adjustments.

A Steadicam in dynamicbalance will pan perfectly on itsown, without constant adjust-ments by the operator. Moreprecise pans and framing arethe result.

Put another way, aSteadicam in dynamic balancewill take full advantage of theexcellent bearings and carefulconstruction of the gimbal.

Dynamic balancefundamentals

Every component on thesled has a mass (or weight) anda position relative to all theother components. The majorand most massive componentsare the battery, the monitor, andthe camera.

Other components includethe electronics, the junctionbox, the Steadicam’s structuralelements, and any accessoriessuch as a small VCR, a receiverfor follow focus control, or avideo transmitter.

Each component has aneffect on both static balanceand dynamic equilibrium.

Each component also has aneffect on the inertial quality, or"feel," of the Steadicam.

These effects can be repre-sented mathematically, and amathematical formula can beused to describe and/or finddynamic balance.

We can also use our under-standing of the math to set up arig and to test for dynamicbalance empirically, withoutsolving any equations.

Depending on theSteadicam model, the operatormay have very few or a lot ofchoices in positioning the majorcomponents.

The operator first positionsthe major components in thebest possible configuration forthe shot, and then the operatorbalances the sled both staticallyand dynamically.

But before we get ahead ofourselves, let’s first look at theforces that affect static balanceand dynamic equilibrium.

3

fig 1a (top view)

common plane

axle

W1

W2

W3

W4

fig 1b (side view)

axlecommon plane

W1

W3

fig 1c (back end view)W

4W

2

If the weights W1 = W

3 and

W2 = W

4 and the distances from

the axle to the weights (theradii) are the same, it will hangvertically and be in dynamicbalance. Sounds logical, butwhy?

Let’s simplify further, andexamine a two mass objectsuspended from the top of theaxle and free to rotate andpivot. Figure 2a essentiallyrepresents a Model I Steadicamwithout a camera, hanging fromits gimbal.

The down arrows representthe force of gravity acting onthe object. If the object is notrotating, then the object willhang perfectly vertical if

(1) W1R1 = W2R2

This is our basic formulafor static balance.

Note: I am numbering theequations in (blue type) tomake it easier to keep track ofthem. Equations in red typeare the most important ones.

The formula for centrifugalforce is

(2) F = MRΩ2

where M is the mass of theobject, R is the radius from thec.g. of the object to the axis ofrotation, and Ω (Omega) is theangular velocity. A commonunit for this is rpm.

The centrifugal forces, F1

and F2, pull the object in the

direction of the arrows.

It is very, very important tounderstand that the centrifugalforces don’t make the objectrotate. The centrifugal forcesare created because of therotation, and these forces growwith the square of the rotationalspeed (Ω).

If the object was in staticbalance (i.e., W

1R

1 = W

2R

2),

then it would be in dynamicbalance if the action of the twocentrifugal forces on the axlewere equal, i.e.,

(3) F1L

1 = F

2L

2

The equation is not F1 = F

2.

How much a centrifugal forcetilts the axle depends upon howfar away from the pivot – orreaction point – the force isapplied to the axle.

Here is a practical way todemonstrate why the distancesto the reaction point (the L

1 and

L2 of figure 2b) matter:

Balance your Steadicamstatically. Take your finger andpoke it at the gimbal, justbelow the bearings. This isclose to the reaction point – andthe force has little effect on therig.

Using the same amount offorce, poke the rig far from thegimbal. Big effect!

Hey, that’s leverage for you,and why we use F times L inour formulas.

4

Also note: The force ofgravity continues to affect theobject as it spins.

If we spin the object, asecond force is created. Thisforce is the centrifugal force.See figure 2b.

fig 2aW2

W1

R1

R2

reaction point

F2

L1

L2

F1

fig 2bM

1M2

R2

R1

We need to assign a con-vention to the way these forcesact on the object.

We will call the forces“positive” if they tilt the object“up” as in figure 2c.

If all the forces that are“positive” equal all the forcesthat are “negative” then theobject won’t tilt as it is rotated,i.e., it will be in dynamicequilibrium. The formula:

(4a) F2L2 + W1R1

= F1L1 + W2R2

Substitute MRΩ2 for F

(4b) (M2R

2Ω2)L

2 + W

1R

1

= (M1R

1Ω2)L

1 + W

2R

2

Because mass and weightare proportional, we can substi-tute weight for mass and get

(4c) (W2R2Ω2)L2 + W1R1

= (W1R1Ω2)L1 + W2R2

We will come back toformula 4c, the basic formulafor dynamic equilibrium, againand again.

Before we move on, take alook at figure 2e.

Now the L’s = zero, andtherefore these centrifugalforces will not tilt the Steadi-cam, even if they were un-equal!!

Mathematically, if L1 and L

2

equal zero, then (4c) becomes

(4d) (W2R

2Ω2)0 + W

1R

1 =

(W1R

1Ω2)0 + W

2R

2

which becomes

(4e) W1R

1 = W

2R

2

In figure 2e, both F’s are inthe same horizontal plane. Theobject would tilt only if thestatic forces, W

1R

1 and W

2R

2,

were unequal.

Now let’s examine anotherobject with 2 masses, but thistime the masses are not copla-nar. Figure 3a essentiallyrepresents a Model III Steadi-cam with a raised monitor andno camera attached.

Note that I’ve also changedthe two radii, just for fun.

5

We will call the forces“negative” if they tilt the object“down,” as in figure 2d.

Note: In the real world, wewould have to run around theobject as it rotated to maintainthe “side view” in all thesediagrams.

We’ve moved the reactionpoint down to the horizontalplane of the two weights.

new reaction point

F1

fig 2e

F2

W2

W1

R1

R2

F2

W1

fig 2c“positive” forces

fig 2d“negative” forces

F1

W2

fig 3a

W2

W1

R1

R2

6

If this object is stationary, itwill remain vertical if the staticforces are equal.

This is our old formula,

(1) W1R

1 = W

2R

2.

Note that static balancedoes not change no matter howmuch W

2 is raised above W

1.

There are no “L’s” in theformula for static balance.

Figure 3b represents thecentrifugal forces acting on thisobject.

Remember F = MRΩ2, andsubstituting W for M, we get

(4c) (W2R

2Ω2)L

2 + W

1R

1 =

(W1R

1Ω2)L

1 + W

2R

2

If it is in static balance,W

1R

1 = W

2R

2.

Thus, equation 4c becomes

(4f) (W2R

2Ω2)L

2 =

(W1R

1Ω2)L

1

Note that all the terms onone side of the equation areequal to the terms on the otherside except L

2 and L

1.

Because L2 is less than L

1,

(W2R

2Ω2)L

2 must be less than

(W1R

1Ω2)L

1, and therefore

equation 4f cannot be satisfied.

This means if our two massobject has its masses on differ-ent horizontal planes, it can notbe in dynamic equilibrium if itis in static balance.

But – thank goodness – athree (or more) mass objectwith masses on different hori-zontal planes, such as aSteadicam – can be balanceddynamically.

It’s time to examine asimple three mass object – anobject similar to a good oldModel III Steadicam with acamera attached.

Look at the next threediagrams. Figure 3c representsall the “positive” forces.

This object will be indynamic equilibrium if

(4a) F2L

2 + W

1R

1

= F1L

1 + W

2R

2

But it can’t be. Why?

F1

fig 3b

F2

fig 3c“positive” forces

W1

F2

Figure 3d represents all the“negative” forces.

fig 3d“negative” forces

F1

W2

Figure 3e represents all theforces and factors relevant todynamic balance.

fig 3e

F2

F1

L2

L1

W2

W1

R1

R2

We will call the masses byname (camera, monitor, andbattery) and look at the forcesaffecting the Steadicam bothstatically and dynamically.

Figure 4a represents oursimple Steadicam and thevarious forces and factors thatwould act upon it.

It is drawn with the camerac.g. to the right of the centralpost, but we don’t really knowyet where to place the camera.

Just as with our two massobject, to be in static balance,the static forces must canceleach other out, i.e., add up tozero. Remember which forcesare “+” and which are “–.”

(5) – WmR

m + W

bR

b

+ WcR

c = 0

Just as with our two massobject, all the forces must addup to zero if it is in dynamicequilibrium.

(6a) FmL

m – W

mR

m

– FbL

b + W

bR

b

– FcL

c + W

cR

c = 0

or, substituting WRΩ2 for F,

(6b) (WmR

mΩ2)L

m – W

mR

m

– (WbR

bΩ2)L

b + W

bR

b

– (WcR

cΩ2)L

c + W

cR

c = 0

That’s a lot of variables!

However, if we choose ourreaction point carefully, we cansolve the equations.

Our first step is to make thereaction point at the samehorizontal plane as the camerac.g., as in figure 4a.

Now, Lc = 0. Therefore,

(WcR

cΩ2)L

c = 0. And if the

object is in static balance,– W

mR

m + W

bR

b + W

cR

c = 0.

So we can remove both“zero” elements from equation6b and get equation 7a.

(7a) (WmR

mΩ2)L

m

– (WbR

bΩ2)L

b = 0

or, expressed another way,

(7b) (WmR

mΩ2)L

m =

(WbR

bΩ2)L

b

Since Ω2 appears in bothterms, we can eliminate that aswell, and we get

(7c) (WmR

m)L

m =

(WbR

b)L

b

Note that equations 7a, 7b,and 7c have no factors thatdepend on the camera weight orradius!!

7

Fig 4a

Lm

Wm

Rm

Wb

Lb

Rb

Rc

Wc

reaction point

Fm

Fb

Fc

Lb

This means that the properplacement of the monitor andthe battery (the terms of equa-tion 7c) will not change, re-gardless of the weight of thecamera. They only depend onL

m and L

b, which are related to

the vertical position of the c.g.of the camera.

We will first use equation7c to find where to place thebattery radius (R

b), and then we

can solve equation 5 to find thecamera radius (R

c).

Note: We could use equa-tion 7c to solve for any of thevariables, but there’s a reasonwe generally use equation 7c tosolve for R

b.

On page 3 we said, “Theoperator first arranges orpositions the major componentsto the best possible positionsfor the shot, and then theoperator balances the sled bothstatically and dynamically.”

So let’s try an example:First we weigh the camera,battery, and monitor, and wedetermine the position of theirrespective c.g.’s.

Next we choose the monitorradius and length for viewingand inertial purposes.

Then we choose the batterylength because we need the sledto be a certain length (forinstance, we want the lens highso we make the sled longer).

To finally put the Steadicaminto dynamic balance, we needto find the battery radius andthe camera radius. We first findR

b by solving equation 7c.

Then we can substitute thevalue of R

b into equation 5, and

we can calculate the requiredvalue of R

c.

When the battery and thecamera are set to these values,the Steadicam will be in perfectdynamic balance.

Remember, the reactionpoint may be moved anywhereon the axis of the post withoutdisturbing dynamic balance.

8

To illustrate, let’s assignsome values. These are valuesthat are fixed, or values that wecan determine by choice.

Values that are fixed

Wc = 25#

Wm = 6#

Wb = 5#

Values that we can chooseto some extent (depends on ourSteadicam, of course)

Lm = 20 inches

Lb = 30 inches

Rm = 10 inches

See figure 4b.

6#

25#

5#

?? Rc

30 inches

20 inches

10 inches

fig 4b

reaction point

?? Rb

9

First, we use equation 7c tofind the battery radius (R

b).

(7c) (WbR

b)L

b = (W

mR

m)L

m

(7d) Rb = (W

mR

m)L

m

(WbL

b)

Rb = (6 x 10) x 20

(5 x 30)

Rb = 8 inches

Now we can use equation 5to find the camera radius (R

c).

(5a) – WmR

m + W

cR

c

+ WbR

b = 0

(5b) WcR

c = W

mR

m – W

bR

b

(5c) Rc = W

mR

m – W

bR

b

Wc

Rc = (6 x 10) – (5 x 8)

25

Rc = 0.8 inches

If we set Rc to 0.8 inches,

and we set Rb to 8 inches, the

Steadicam will be in both staticbalance and dynamic equilib-rium, i.e., in dynamic balance.

Now that we’ve solvedequations 7c and 5, we candemonstrate that moving thereaction point has no effect ondynamic balance.

As an example, let's movethe reaction point six incheslower down the post, to where agimbal might be placed. Seefigure 4c.

We start by restating theequation 6b, our formula fordynamic equilibrium:

(6b) (WmR

mΩ2)L

m – W

mR

m

– (WbR

bΩ2)L

b + W

bR

b

– (WcR

cΩ2)L

c + W

cR

c = 0

Again, if it is in staticbalance, the static terms equalzero. The Ω2 appears in eachremaining term, so we caneliminate Ω2, and we are leftwith equation 8.

(8) (WmR

m)L

m – (W

cR

c)L

c

– (WbR

b)L

b = 0

Dropping the reactionpoint six inches means that now

Lm = 14, L

c = minus 6, and

Lb = 24

Substituting these valuesinto equation 8, we get

(6 x 10) x 14 – (25 x 0.8) x (– 6)

– (5 x 8) x 24 = 0

or

840 + 120 – 960 = 0

Which demonstrates thatthe Steadicam will remain inperfect dynamic balance,regardless of where we placethe gimbal on the central post.

fig 4c

5#

6#

25#

0.8 inches

24 inches

14 inches 10 inches

new reaction point

8 inches

6 inches

10

We have now discussed allthe basic physics and math-ematics that we need to under-stand dynamic balance.

But if we want a bettermathematical model, all themajor masses should be ac-counted for in the equations.

The Steadicam of figure 4cis only a three mass object. Wehave not accounted for theelectronics, or the J-box, or thestructure of the Steadicam, orany accessories we might addto the sled.

Each of these masses has astatic effect and, when theSteadicam is rotated, each masscreates a centrifugal force. Infigure 5a, we are illustratingwhat happens when just twomore masses are added to thestructure.

In our convention of plusand minus forces, a massforward of the post (like themonitor) will have a positivecentrifugal force and a negativestatic force. A mass added tothe rear (like the battery) willhave a negative centrifugalforce and a positive static force.

As depicted in figure 5a, thefocus motor receiver has apositive F

f and a negative W

fR

f.

The electronics package has anegative F

e and a positive W

eR

e.

Each force must be addedinto the two balance equations.

Reworking our staticformula (5) to account for thefocus receiver and the electron-ics, we now get

(9) – WmR

m + W

bR

b +

WcR

c – W

fR

f + W

eR

e = 0

and our dynamic equilib-rium formula (6a) becomes

(10) FmL

m – W

mR

m – F

bL

b

+ WbR

b – F

cL

c + W

cR

c + F

fL

f –

WfR

f – F

eL

e + W

eR

e = 0

We would solve theseequations in the same manneras before, by starting withequation 10 and assuming thesled is in static balance. Wewould then remove the staticelements and substitute MRΩ2

for F, and W for M.

Again we would solve theequations for the proper batteryradius, and then use that valuein equation 9 to solve for thecamera radius.

If we want to sufficientlymodel the Steadicam, weshould account for the J-box ofthe Steadicam, significantstructural components, andevery accessory we might addor remove. For instance, theUltra’s computer programaccounts for five major compo-nents, any camera, and threeaccessories.

It’s clear that solving theseequations quickly becomes anightmare, unless one has acomputer program.

Le

We

Fe

Ff

Wf

Rf

Re

Lf

reaction point

Fig 5a

If we have a sled thatallows us many choices in theconfiguration of major compo-nents, we can use the math tomake a setup chart for variousmodes – long, short, mediumlength, monitor high or low,monitor extended, with andwithout a VCR attached, etc.We can use the chart to quicklyget the sled very close todynamic balance.

If we have a sled withrelatively few choices in theconfiguration of major compo-nents, we can use the math todiscover how to add accessoriesin a manner that will continueto make it possible to achievedynamic balance.

If we have some sort ofcustomized hybrid conglomera-tion of parts, we can check tosee if this sled can get intodynamic balance in all itsconfigurations, or if we shouldgo back to the shop and makesome new parts.

We can use the math tounderstand why dynamicbalance in low mode can be justas easy to achieve as dynamicbalance in high mode.

And that’s just a start....

Without measuring orweighing anything, or solvingany equations, our under-standing of the math canmake our life – both on andoff set – much, much easier.

If we understand the basictheory, we can quickly andeasily position the sled compo-nents very closely to a condi-tion of good dynamic balance.

We also can use our knowl-edge to closely predict whatchanges we need to make if wemove a major component, or ifwe add an accessory, changelenses, or tilt the monitor, etc.

We can use the theory toknow that after moving acomponent, all we ever have todo to get back into dynamicbalance is to move the batteryin or out a bit, and then staticbalance with the camera.

We can use the theory toknow how to add or placecomponents on the sled whereit doesn’t make it difficult orimpossible to get the sled intodynamic balance.

We can use the math tounderstand why a given camerawill always be placed in a verynarrow range of positions.

We can use the math toavoid tests and procedures thatare irrelevant or wasteful of ourtime and energy.

11

Some practical limitations

Our mathematical model israther simple, and doesn’taccount for many things,including wind effects, or theuneven distribution of mass in agiven object. For instance, notall internal parts of a monitorhave the same density, so themonitor doesn’t behave exactlylike a tiny, very dense mass atthe c.g.

The key thing to rememberis that our formulas are only amodel of reality.

So why did we just gothrough all that math?

Because the math clearlydescribes the major forcesaffecting dynamic balance.

Because over the last 15years, this model has beenproven to be an extremelyuseful tool. In spite of itssimplicity, the model worksextremely well in practice,where it all matters.

For instance, using the math(and a Palm Pilot™ computerto solve the equations), one cangenerally place the batterywithin .25 inch of the idealposition. A quick spin test canfine tune the system to theneeded degree of precision.

But that’s only one minorreason that the math and theoryare important.

Mount the camera to thedovetail, and add all lenses,mags, film, motors, etc., to thecamera. Mark the fore-aft c.g.position.

Position all the componentson the sled, making sure theyall line up in a single verticalplane fore and aft, i.e., don’tadd components to the side ofthe sled.

If you have a choice, makethe sled the length you want.Position the monitor high orlow, and/or in or out, to thebest possible advantage for theshot.

12

A simple,empirical method

to getany Steadicam

intodynamic balance

Mount the camera to thestage, and slide the camera sothat the camera c.g. is about .75inch to the rear of the centerlineof the central post.

Place the gimbal on thebalancing stud of your standand balance top to bottom sothat you have a relatively longdrop time, say 3 to 4 seconds.

Balance precisely side toside by moving the camera.

Balance fore and aft as bestas you can using the battery.This is very important.

Now fine tune the fore andaft balance using the camera.

Be sure to get the rig intovery good static balance, bothside to side and fore and aft.

Give it a few spins, but nottoo fast a spin, say 20 to 40 rpmto start. If you are close todynamic balance, you can spinit as fast as you want.

If it’s not in acceptabledynamic balance, move thebattery in or out about .25 inch.Static balance with the cameraand try again. Is it better orworse? Change it again untilyour rig pans consistently onthe axis of the post.

Done.

A good tip: make a note ofthe camera radius, and theprocedure will be even fasterthe next time.

This whole balancingprocess should take about twominutes. Do not spin the cam-era at excessive speeds and donot fuss too much.

If you change lenses, addfocus motors, etc. to the cam-era, the sled will be out ofdynamic balance, but usuallyonly very, very slightly. Mov-ing the camera fore and aft torestore static balance will keepyou in good dynamic balance.Why?

The placement of thebattery only depends on thevertical distance of the compo-nents to the c.g. of the camera(as in formula 7d), not on thecamera’s weight or radius.

(7d) Rb = (W

mR

m)L

m

(WbL

b)

Changing lenses or addingfocus motors, glass filters, etc.usually doesn’t change thevertical c.g. of the camera verymuch; therefore it will notaffect the proper placement ofthe battery very much either.

We can easily determinewhere to place the “alteredcamera” by static balancing onthe stand. Note that the mathwe would use to determine thecamera radius, formula 5c, is anexpression of static balance.

(5c) Rc = W

mR

m – W

bR

b

Wc

13

Now let’s look at a simple,non-mathematical model forunderstanding dynamicbalance.

Figure 6a looks like theModel I or the SK. The camerac.g. is centered over the post;the monitor and battery are onthe same horizontal plane, andtheir common c.g. – the red dot– is under the post. This unit isin dynamic balance.

Figure 6b looks like aModel III in dynamic balance.The battery c.g. is closer to thepost, and the camera c.g. hasmoved to the rear.

Why?? See the third figure.

Fig 6a

In figure 6b, the monitorradius remains the same, butthe monitor raised a bit.

In figure 6c, the monitorhas been raised all the way upin front of the camera. Thecommon monitor and camerac.g. – another red dot – is nowover the post, the battery’s c.g.is under the post, and thisSteadicam also is in dynamicbalance.

Figure 6c is an absurdarrangement, of course, but itmakes a point. As the monitoris raised, the camera c.g. mustmove to the rear and the batteryc.g. must move towards thepost.

A rule of thumb: If themonitor is raised about 25 to 30percent of the way up from thebattery level, the camera c.g.will be about .75 inch behindthe centerline of the centralpost, more or less.

Fig 6c

Fig 6b

If we shorten the rig (orjust raise the monitor), themonitor is raised a greaterpercentage of the total distancefrom the battery to the camera.Therefore the battery will bemoved forward, and the camerafurther to the rear, as it is infigure 6e.

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What to do if we make thesled longer or shorter?

Assume we have a sled indynamic balance. If we extendthe sled by moving both thebattery and monitor away fromthe camera, we create a newsituation. It is useful to think interms of how much – percent-age wise – the monitor is raisedabove the battery.

As in figure 6d, the moni-tor is raised by a smallerpercentage of the total distancefrom the battery to the camerathan before. Therefore, thebattery will be more to the rear,and the camera moved forward.

What to do if we addaccessories?

Our simple method ofgetting a rig into dynamicbalance, the non-mathematicalconceptual model on page 13,and our mathematical model allattempt to place the battery inthe right position as the firststep to get a rig into dynamicbalance.

Therefore, when I add anaccessory, I like to think of howthe placement of the accessorywill change how much “work”the battery must do to achievedynamic balance.

Thus, if an accessory isadded in front of the post, thebattery must “work harder” tocounteract this mass and itsforces, and I would move thebattery more to the rear. Seefigure 6f. Also, the lower theaccessory is placed, the harderthe battery has to work.

Fig 6e

If an accessory is addedbehind the post, as in figure 6g,it “helps” the battery do itswork.

Therefore I would movethe battery forward before Iwould static balance with thecamera and spin test for dy-namic balance.

Fig 6f

Fig 6g

Fig 6d

15

Using a dynamic balanceprogram

Some things not to doif you want your sled to be in

dynamic balance

Do not add components tothe side of the rig, and do notbalance side to side usinganything other than the cameramounting stage. Why?

Take another look at figure7a (the same figure as 1c).

The components W2 and W

4

can represent accessoriesmounted on each side of a sled.The forces they generate act totilt the post sideways, not foreand aft like all the others wehave been discussing. W

2 and

W4 create an entire new set of

forces that must also be equal,just as the “fore and aft” forceshave to be equal for dynamicbalance.

While it is possible to dothe math (with a computer, ofcourse) to determine where toposition the battery and camera,it is quite difficult to determineempirically what to do. Shouldone move the battery – and/orthe camera – side to side tocorrect a problem, or fore andaft, or some combination of thetwo? There is no way to know.

Worse yet – for dynamicbalance – is to balance side toside with the battery. Assumingthe other rig components arenicely organized fore and aft, ifthe operator balances side toside with the battery, it meansthat the camera c.g. was dis-placed to the other side forstatic balance. See figure 7b.

fig 7a (back end view)

W4

W2

fig 7b (back end view)

We have created a situationidentical to that of figures 3athrough 3e. We’ve alreadyshown that this arrangement ofmasses cannot be in dynamicequilibrium if it is in staticbalance, i.e., it can not be indynamic balance. (See pages 5and 6.)

The Model I, II, and IIISteadicams “wagged” theirbatteries side to side. If one waslucky, the camera was mountedon the dovetail so that thecamera’s c.g. was close to theaxis. But from the perspectiveof dynamic balance, it wasn’t agood design.

A computer program fordynamic balance is a greateducational tool. It will showexactly what happens to thebattery and camera radii whenany specific change is made tothe sled. It will demonstratewhich changes make a lot ofdifference, and which do not.

A good dynamic balanceprogram can also help anoperator get the rig into dy-namic balance quickly, regard-less of the many possibleconfigurations of the rig. Agood program accounts for allthe major masses of the Steadi-cam, any camera, and for manyaccessories.

As the Steadicam isreconfigured on set, the opera-tor only has to take three newmeasurements (at most), andthe computer determines theexact placement of the batteryand the camera. Spin balancingis generally unnecessary, whichis really great if you flip to lowmode and lengthen the sled.

A static test for dynamicbalance?

Alas, it can not be done.The centrifugal forces are notcreated until the rig is rotated.All static tests – no matter howfinely conducted – can onlyaccount for static forces. If youdon’t spin it, the centrifugalforces don’t exist. How else canI say it? For instance, if you laythe sled horizontally, the test(fig 8) is exactly like the statictest of figure 2e. (See page 5.)

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the sled, or only pairs of com-ponents carefully balanced onthe same horizontal plane. Thesimple procedure on page 12,however, works with anySteadicam in any configuration.

Trimming for headroom

Operators should alwaysuse the balance of the Steadi-cam to get a shot.

Often we deliberately movethe camera to set the sled atsome angle other than vertical.If the shot has only slow pans,the centrifugal forces are small,and it may be far more impor-tant to have the sled hang at agiven angle than to be indynamic balance.

On the other hand, try thistest. Get your rig into perfectdynamic balance on the stand.Then move the camera forwardto tilt the sled (and the camera)a few degrees and give it agood spin. What happens? Therig behaves very strangely, veryquickly.

Tilting the monitor

Try this test: Get the riginto perfect dynamic balance onthe stand, then tilt the monitorand spin it again.

Unless the monitor tilts onits c.g., both the L

m and the R

m

of the equations have changed,and the sled will no longer bein dynamic balance.

If a static test “works,” it isonly because the rig is in aunique configuration, with themonitor and battery on thesame horizontal plane.

Additionally there must beeither no other components on

If you want to angle thecamera up or down and alsoremain in dynamic balance, youneed either an integral tilt heador a wedge plate. See the UltraManual for more informationabout an integral tilt head.

fig 9tilting a Model III monitor

Wb

Rm

Rb

Wm

reaction point

fig 8 (back end vieww/ horizontal post)

new reaction point

F1

fig 2e(side view)

F2

W2

W1

R1

R2

Try the test with the sledboth very long and short. Whathappens now?

This monitor tilts close toits c.g., so that tilting it has verylittle effect on the dynamicbalance of the sled.

A couple of extra things

Here is a quick and easymethod (no math!) for gettingyour sled into dynamic balancewhen the monitor is displacedsideways, or you want thecamera facing a direction otherthan forward.

The conceptual trick is tothink in terms of all the c.g.’s –the black dots of our diagrams.In our model, the orientation ofany particular part does notmatter, nor does it matter whatshape it is. We are only inter-ested in the mass of the objectand the position of the c.g.

17

Conclusion

Dynamic balance is impor-tant for precise operating.

A solid conceptual under-standing of dynamic balancewill help the operator get anySteadicam into dynamic bal-ance quickly and efficiently,without doing any math.

It will also make it easy toget the Steadicam back intodynamic balance if any changesare made to the configuration.

Understanding dynamicbalance – what it is, what it isnot, and its practical limitations– will also help the operatordecide how important dynamicbalance is in any situation.

Understanding and usingthe math not only helps concep-tually, but, with the aid of ahandheld computer, it can be anextremely useful tool on set.

The math is also crucial forthose who modify their equip-ment and want to be able to gettheir rigs in dynamic balance inall possible configurations.

It is my hope that thisdocument will help operatorseverywhere understand thisimportant subject.

Jerry Holway,April 1st, 2003

All one has to do is to keepthe c.g.’s in one vertical plane,(as we always should do) andthe Steadicam easily can be putinto dynamic balance.

Which is to say, keep thebattery directly behind themonitor, and aim the camera inany direction you choose. Startby placing the camera c.g.about .75 inch behind the post(“behind” the post is always inthe vertical plane of the batteryand monitor). Balance asdescribed on page 12.

It’s easier to operate if themonitor and camera face in thesame direction or exactly 180˚from each other.

A fun thing to do

Put your sled into perfectdynamic balance, then place thegimbal so that the sled isneutrally balanced top tobottom. Now the sled will hangat any angle and still spinconsistently on the axis of thecentral post.

This is even more fun if thesled is very long, as you can setthe rig panning and also gyratethe whole thing in huge arcs.

You can move throughspace at the same time, and theSteadicam becomes a reallyelegant and magical object.