AMB Design Spreadsheet Equations

Ari Meles-Braverman

Version 1.3

Contents

1 General Motor Equations 2

2 Mechanism Gear Ratio Calculator 3

3 Gear Options Calculator 5

4 Chain/Belt C-C Calculator 5

5 Drivetrain Calculator 6

6 Projectile Trajectory Calculator 7

7 Pneumatics Calculator 9

8 Pneumatic Linkage Calculator 10

9 Lead Screw Calculator 11

10 Beam Bend Calculator 12

Appendix A Maximum Efficiency Derivation 16

Appendix B Chain Length Derivation 17

Appendix C Optimal Shifting Speed Derivation 18

Appendix D Projectile Equation Derivations 19

Appendix E Pneumatics Simulation Derivation 20

Appendix F Equivalent Radius/Load Derivation 21

1

1 General Motor Equations

1.1 Motor System Equations

x denotes the property of a system of equivalent motors connected 1:1 in a gearbox

ωf = ωf ·(

V

Vspec

)(1)

τs = τs · ηn(

V

Vspec

)(2)

if = if · n(

V

Vspec

)(3)

is = is · n(

V

Vspec

)(4)

P =2πωf · τs

4(5)

P =2πωf · τs

4= P · ηn

(V

Vspec

)2

(6)

KT =τs

is − if(7)

ωf = Motor Free Speedωf = Adjusted Free Speedτs = Motor Stall Torqueτs = Adjusted Stall Torqueif = Motor Free Currentif = Adjusted Free Currentis = Motor Stall Currentis = Adjusted Stall CurrentP = Motor PowerP = Adjusted Motor Powern = # of Motorsη = Gearbox EfficiencyV = Applied VoltageVspec = Specification Voltage (Almost Always 12)KT = Motor Torque Constant

1.2 Instantaneous Motor Equations

ω = ωf · (1−τ

τs) (8)

i = (is − if )τ

τs+ if (9)

ηmotor =Wout

Win=τ · ω(τ)

V · i(τ)=

τ(τs − τ)ωf

V (isτ + if (τs − τ))(10)

ω = Instantaneous Motor Speedτ = Instantaneous Motor Torquei = Instantaneous Motor Currentηmotor = Instantaneous Motor Efficiency

2

2 Mechanism Gear Ratio Calculator

2.1 General Equations

ωfree =ωfG

(11)

Fs =τsG

r(12)

ωload = ωfree · (1−F

Fs) (13)

vfree = ωfree · 2πr (14)

vload = ωload · 2πr (15)

i =rF

KTGn+ifn

(16)

Vs = VspecrF

τsηnG(17)

G = Gear Ratioη = Gearbox Efficiencyn = # of MotorsF = Load Appliedr = Load Radiusωfree = Output Rotational Free Speedωload = Output Rotational Loaded Speedvfree = Output Linear Free Speedvload = Output Linear Loaded Speedi = Current Per MotorFs = Stall LoadVs = Stall Voltage

2.2 Gear Ratio Calculations

2.2.1 Maximum Power

Fs = 2F (18)

Substituting into (12) and solving for G gives:

G =2rF

τs(19)

2.2.2 Maximum Efficiency

G =rF

τs

(1 +

√is

if

)(20)

For derivation, see Appendix A

3

2.2.3 At Stall

Fs = F (21)

Substituting into (12) and solving for G gives:

G =rF

τs(22)

2.2.4 By Rotational Speed (ωload)

We can substitute (8) and (9) into (10):

ωload =ωfG· (1− F

τsGr

) (23)

Solving for G, we get:

G =ωf

2ωload

(1 +

√1− 4rF

ωloadτsωf

)(24)

2.2.5 By Linear Speed (vload)

We can solve (15) for ωload:

ωload =vload2πr

(25)

Now knowing ωload we can use (24) to calculate G

2.2.6 By Per-Motor Current (i)

Solving (16) for G, we get:

G =rF

KT (ni− if )(26)

2.2.7 By Stall Load (Fs)

We can solve (12) for G:

G =rFsτs

(27)

2.2.8 By Stall Voltage (Vs)

Solving (17) for G, we get:

G = 12rF

τsηnVs(28)

4

3 Gear Options Calculator

This calculator uses a brute-force search algorithm to find combinations of gears thatmeet the set criteria. All gears in the list must have dp = 20. The formulas used tocheck the criteria are:

ODx =nx + 2

20[in] (29)

C-Cxy =nx + ny

40[in] (30)

clearancexyz = C-Cxy − 12ODz (31)

nx = Number of teeth of gear x ∈ [1, 2, 3, 4]G = Total gear ratioODx = Outer diameter of gear xC-Cxy = Center-to-center distance between gears x and yclearancexyz = Clearance distance of gear x shaft

4 Chain/Belt C-C Calculator

This method of solving for D is inspired by and based on the formulas used in Clem1640’sChain/Belt Calculator. The language in this section will only reference chain, sprockets,and chain links; but all equations also apply to belts, pulleys, and belt teeth, respectively.

First we start with basic equations for chain and sprockets:

2πri = nip (32)

L = lp (33)

C = D − (r1 + r2) (34)

We can construct an equation to represent the path length of the chain relative to thecenter-to-center distance. A derivation of this equation can be found in Appendix B.

L = 2√D2 − (r1 − r2)2 + 2(r1 − r2) cos−1

(r1 − r2

D

)+ 2πr2 (35)

Unfortunately we cannot solve this equation analytically to get D, even with the help ofa computer. So we solve the function numerically using the Newton-Raphson Method.For the forward calculation our initial guess for D is L

2 , which would be the case if bothsprockets had diameters of 0. For the reverse calculation, we use the estimated value ofD as the initial guess. The Newton-Raphson Method then improves the accuracy of Dto the correct solution until the error is less than 10−12.

L = Chain Lengthl = # of LinksD = Center to Center Distanceni = # of Teeth on Sprocket i ∈ [1, 2]ri = Radius of Sprocket i ∈ [1, 2]p = Chain PitchC = Clearance

5

5 Drivetrain Calculator

5.1 Forward Calculation

Equations taken from JVN’s Mechanical Design Calculator.

vfree =ωf · πdG

(36)

vadj = vfree · kSL (37)

islip =Wµd

2KT · nG+ifn

(38)

5.2 Reverse Calculation

itot =Vbatt − Vmin

Rbatt +Rmain + 1nRbr

(39)

Gslip =µdW

2KT

(itot − if

) (40)

Formulas for vfree and vadj are the same as for the Forward Calculation – (36) and (37)respectively

vfree = Free Speedvadj = Adjusted Speedislip = Wheel Slip Currentitot = Maximum Total Current DrawGslip = Maximum Gear Ratio to Slip Wheels above VminKT = Motor Torque Constantωf = Motor Free Speedif = Motor Free Currentd = Wheel Diameterµ = Wheel Coefficient of FrictionW = Adjusted Robot Weight (i.e. weight resting on driven wheels)G = Total Gear RatiokSL = Speed Loss ConstantVbatt = Battery Resting VoltageVmin = Minimum Allowable System VoltageRbatt = Battery Internal ResistanceRmain = Resistance of Main Power WiringRbr = Resistance of Each Branch Circuit

6

6 Projectile Trajectory Calculator

6.1 Forward Calculation

The Forward Calculation uses equations taken from this paper by the University ofIllinois. Three forces can be identified as acting on the ball: gravity, Fg, air drag, FD,and lift, FL. The acceleration from these forces is applied at each timestep until the ballreaches the target or hits the ground.

Fg = mg (−y) (41)

FD =1

2ρ|v|2AcCD (−v) =

π

2ρr2v2CD (−v) (42)

FL =1

2ρ|v|2AcCL (ω × v) =

π

2ρr2v2CL (ω × v) (43)

Combining these gives ∑Fx = max = −FL sin θv − FD cos θv

= −π2ρr2v2(CL sin θv + CD cos θv)

= −π2ρr2v(CLvy + CDvx)

(44)

∑Fy = may = FL cos θv − FD sin θv − Fg

=π

2ρr2v2(CL cos θv − CD sin θv)−mg

=π

2ρr2v(CLvx − CDvy)−mg

(45)

To find the drag coefficient we use an empirically determined graph correlating the dragcoefficient with the Reynolds number, Re = ρvDµ−1. Since this is a strongly non-linearcorrelation, we will use an estimated average velocity to calculate an approximate dragcoefficient to be used throughout the trajectory.

To find the lift coefficient we use the spin factor, S = rωv−1. The Illinois paper finds astrong correlation between S and CL, with CL = 1.6S for S < 0.1 and CL = 0.6S + 0.1for S > 0.1. At every timestep, S and therefore CL are recalculated based on the up-dated velocity v.

CD = Drag CoefficientCL = Lift Coefficientρ = Air Densityr = Ball Radiusv = Ball Velocityω = Ball Rotational Velocitym = Ball Massg = Gravity Constantθv = Velocity Angle (from +x)

7

These equations are good approximations of drag and lift for spherical rotating balls.For any object in flight you can use the following traditional trajectory equations tocalculate the object’s path with negligible air drag and lift. These will give the sameoutputs as the simulation equations above when CD and ω are set to 0.

hf = hi + d tan θi −g · d2

2v2i cos2 θi

(46)

θf = tan−1

(2hf − hi

d− tan θi

)(47)

6.2 Reverse Calculation

θi = tan−1

(2hf − hi

d− tan θf

)(48)

vi = sec θi ·

√g · d

|tan θi − tan θf |(49)

hi = Release (Initial) Heighthf = Target (Final) Heightθi = Release (Initial) Angleθf = Target (Final) Anglevi = Release (Initial) Velocityd = Horizontal Distance to Targetg = Acceleration Due to Gravity

For derivation see Appendix D.

8

7 Pneumatics Calculator

The ”simulation” is run at timesteps of dt = 1 second for a duration of 150 seconds.

The pressure at the current step of the simulation can be calculated by:

Pn+1 = Pn +Wcomp −

∑Wcyl

V(50)

Wcomp = Vcomp(P ) · dt · 1atm (51)

Wcyl =

(πD2

4Ppush +

π(D2 − d2)

4Ppull

)L ·m (52)

Vcomp(P ) is the compressor flow-rate as a function of the output pressure, taken froma fourth-degree polynomial interpolation of the data provided online. The volume ofair is measured at atmospheric pressure, not at system pressure. m is the number ofactuations per second (can be 0 when not firing, 1 when firing, or m > 1 if firing morethan once per second).

Derivations of these equations can be found in Appendix E.

We can also calculate the pushing and pulling force of each cylinder using the formulas:

Fpush =πD2

4Ppush (53)

Fpull =π(D2 − d2)

4Ppull (54)

Pn = System Pressure at Step nV = System VolumeWcomp = Work Done by the CompressorWcyl = Work Done by Each CylinderD = Cylinder Diameterd = Cylinder Rod DiameterL = Cylinder LengthPpush = Pushing PressurePpull = Pulling PressureFpush = Pushing ForceFpull = Pulling Force

9

8 Pneumatic Linkage Calculator

Let θ1 and α1 represent the system when the cylinder is retracted, and θ2 and α2 rep-resent it when extended.

In the horizontal direction we have:

A+X cos θ1 − Y sin θ1 = L cosα1 (55)

A+X cos θ2 − Y sin θ2 = (L+ ∆L) cosα2 (56)

And in the vertical direction we have:

X sin θ1 + Y cos θ1 = B − L sinα1 (57)

X sin θ2 + Y cos θ2 = B − (L+ ∆L) sinα2 (58)

We can combine (55) and (57) to remove α1 and get:

L2 = A2 +B2 +X2 + Y 2 + 2(AX −BY ) cos θ1 − 2(BX +AY ) sin θ1 (59)

And combine (56) and (58) to remove α2 and get:

(L+ ∆L)2 = A2 +B2 +X2 + Y 2 + 2(AX −BY ) cos θ2 − 2(BX +AY ) sin θ2 (60)

To derive the forward formulas, you can solve (59) and (60) separately to get expressionsfor θ1 and θ2. To derive the reverse formulas, you can solve (59) and (60) as a systemof equations to get expressions for A and B. I will not include these expressions in thisdocument, because they are extremely long.

10

9 Lead Screw Calculator

The basic transformation between rotational and linear motion is given by:

v = ω · np (61)

Force & torque equations are taken from Shigley’s Mechanical Engineering §8-2

dm = d− p

2(62)

TR =Fdm

2

(πµdp + np cos(α)

πdp cos(α)− µnp

)(63)

TL =Fdm

2

(πµdp − np cos(α)

πdp cos(α) + µnp

)(64)

Backdrive-able if TL < 0.

η =TR|µ=0

TR=

np

πdp· πdp cos(α)− µnpπµdp + np cos(α)

(65)

Equivalent radius/load formulas are designed to allow for lead screws as outputs in theMechanism Gear Ratio Calculator. Derivations can be found in Appendix F.

req =np

2π(66)

LR,eq =TRreq

(67)

LL,eq =TLreq

(68)

d = Screw Diameterp = Screw Pitchn = # of Startsα = Half Thread Angle (i.e. thread angle

2 )µ = Coefficient of Friction of ScrewF = Applied Forcev = Instantaneous Linear Speedω = Instantaneous Rotational Speeddm = Mean (Average) DiameterTR = Raise TorqueTL = Lower Torqueη = Efficiencyreq = Equivalent RadiusLR,eq = Equivalent Raise LoadLL,eq = Equivalent Lower Loadvfree = Free Speedvadj = Adjusted Speedislip = Wheel Slip Currentitot = Maximum Total Current DrawGslip = Maximum Gear Ratio to Slip Wheels above Vmin

11

KT = Motor Torque Constantωf = Motor Free Speedif = Motor Free Currentd = Wheel Diameterµ = Wheel Coefficient of FrictionW = Adjusted Robot Weight (i.e. weight resting on driven wheels)G = Total Gear RatiokSL = Speed Loss Constant

vshift =v1v2

v1 + v2(69)

vshift = Optimal Shifting Speedv2 = Gear 1 Free Speedv2 = Gear 2 Free Speed

For derivation see Appendix C.

9.1 Reverse Calculation

itot =Vbatt − Vmin

Rbatt +Rmain + 1nRbr

(70)

Gslip =µdW

2KT

(itot − if

) (71)

Formulas for vfree and vadj are the same as for the Forward Calculation – (36) and (37)respectively

Vbatt = Battery Resting VoltageVmin = Minimum Allowable System VoltageRbatt = Battery Internal ResistanceRmain = Resistance of Main Power WiringRbr = Resistance of Each Branch Circuit

10 Beam Bend Calculator

10.1 Geometrical Deformation Resistance

Formulas for deformation resistance constants of different cross-sectional geometries aretaken from StructX.

I = Area (Second) Moment of InertiaJ = Torsion Constant

10.1.1 Hex

I = 0.0601a4 (72)

J = 0.1154a4 (73)

a = Distance Between Flats

12

10.1.2 Round

I =π

64d4 (74)

J =π

32d4 (75)

d = Diameter

10.1.3 Round Tube

I =π

8d3h (76)

J =π

4d3h (77)

d = Diameterh = Wall Thickness

10.1.4 Square

I =a4

12(78)

J =9

64a4 (79)

a = Side Length

10.1.5 Rectangular Tube

I =1

3x2yh (80)

J =2h2(x− h)2(y − h)2

h(x+ y − h)(81)

x = Length Parallel to Forcey = Length Perpendicular to Forceh = Wall Thickness

10.2 Displacement Equations

Displacement formulas are taken from Shigley’s Mechanical Engineering Appendix A-9

10.2.1 Force Between Supports

13

y =Fx2b2

6EIl3[3al − x(3a+ b)] (82)

Substitute b = l − a to get:

y =Fx2(a− l)2

6EIl3[3al − x(2a+ l)] (83)

Take the derivative, set equal to zero, and solve for x:

0 =∂y

∂x=Fx(l − a)2

2EIl3(2a(l − x)− lx) (84)

x =2al

2a+ l(85)

Substitute (85) into (83) to get:

ymax =2Fa3(l − a)2

3EI(2a+ l)2(86)

y = Vertical DisplacementF = Load Forcea = Distance to Closer Supportl = Distance Between SupportsE = Modulus of ElasticityI = Area Moment of Inertia

10.2.2 Cantilevered Force

ymax =Fa2

6EI(a− 3l) (87)

ymax = Largest DisplacementF = Load Forcea = Distance to Supportl = Total LengthE = Modulus of ElasticityI = Area Moment of Inertia

14

10.3 Buckling Force

This formula is taken from the Wikipedia page on Euler’s Critical Load

Fmax =π2EI

(KL)2(88)

Fmax = Maximum ForceE = Modulus of ElasticityI = Area Moment of InertiaK = End Condition ConstantL = Column Length

10.4 Twisting Torque

This formula is taken from Shigley’s Mechanical Engineering §3-12

θmax =TL

GJ(89)

θmax = Largest Angular DisplacementT = Applied TorqueL = Distance Between Torque and SupportG = Shear ModulusJ = Torsion Constant

15

Derivations

A Maximum Efficiency Derivation

To find the maximum efficiency, we take the derivative of (10) and set it equal to 0:

0 =∂η

∂τ=

(if (τ − τs)2 − isτ2)ωf

V (isτ + if (τs − τ))2(90)

Solving for the value of τ at which η is maximized:

τηmax =τs

√if√

if +√is

(91)

We know that τmanip = G · τmotor, so we can substitute τηmax for τmotor:

G =τmanipτmotor

=r · Fτηmax

=r · Fτs√if√

if+√is

(92)

And simplifying gives:

G =rF

τs

(1 +

√is

if

)(20)

16

B Chain Length Derivation

It is clear that:L = 2X + r1 · 2β + r2 · (2π − 2β) (93)

Looking at the right triangle in the center of the image, we can see that:

D2 = X2 + (r2 − r1)2 =⇒ X =√D2 − (r1 − r2)2 (94)

β = cos−1

(r2 − r1

D

)= cos−1

(r1 − r2

D

)(95)

Substituting (94) and (95) into (93) gives:

L = 2√D2 − (r1 − r2)2 + r1 · 2 cos−1

(r1 − r2

D

)+ r2 ·

(2π − 2 cos−1

(r1 − r2

D

))(96)

And rearranging gives:

L = 2√D2 − (r1 − r2)2 + 2(r1 − r2) cos−1

(r1 − r2

D

)+ 2πr2 (35)

17

C Optimal Shifting Speed Derivation

The optimal shifting speed is the speed where both high and low gears produce the sameforce at the wheel surface.

We will start with the definition of instantaneous motor speed from (8). This can besolved for the motor torque τ

τ = τs

(1− ω

ωf

)(97)

Using the equations,

vw =ωr

G(98)

Fw =τG

r(99)

we can transform the rotational speed and torque at the motor into the linear speed andforce at the wheel

Fw =τsG

r

(1− vwG

rωf

)(100)

To find the optimal shifting speed, we will set the wheel forces at both gear ratios, G1

and G2, and both wheel radii, r1 and r2, equal at the shifting speed vshift

τsG1

r1

(1−

vshift ·G1

r1ωf

)=τsG2

r2

(1−

vshift ·G2

r2ωf

)(101)

This can then be simplified to

vshift =r1r2ωf

G2r1 +G1r2(102)

Plugging in ω = ωf into (98) gives the free speeds v1 and v2 for each gear ratio

v1 =ωfr1

G1

v2 =ωfr2

G2

(103)

Then it is easy to see that

vshift =1

1v1

+ 1v2

=v1v2

v1 + v2= v1 ‖ v2 (69)

18

D Projectile Equation Derivations

Starting with basic projectile motion equations:

∆x = vi cos θi · t (104)

∆y = vi sin θi · t− 12gt

2 (105)

θf = tan−1

(vf,yvf,x

)= tan−1

(vi sin θi − gtvi cos θi

)(106)

Solving (104) for t and substituting into (105) gives:

∆y = ∆x tan θi −g · (∆x)2

2v2i cos2 θi

(107)

We can then substitute in ∆x = d and ∆y = hf − hi:

hf = hi + d tan θi −g · d2

2v2i cos2 θi

(46)

Solving (104) for t and substituting into (106) gives:

θf = tan−1

(vi sin θi − g · ∆x

vi cos θi

vi cos θi

)= tan−1

(tan θi −

g · dv2i cos2 θi

)(108)

Further substituting in (46) into (108) gives:

θf = tan−1

(tan θi −

2

d(d tan θi − (hf − hi))

)(109)

And by simplifying we get:

θf = tan−1

(2hf − hi

d− tan θi

)(47)

To calculate the reverse equations, we can solve (109) for θi:

θi = tan−1

(2hf − hi

d− tan θf

)(48)

Then substitute (46) into (48):

tan θi + tan θf =2

d

(d tan θi −

g · d2

2v2i cos2 θi

)(110)

And solve that for vi:

vi = sec θi ·

√g · d

|tan θi − tan θf |(49)

19

E Pneumatics Simulation Derivation

We know that the energy of the system at any point can be represented by:

E = P · V (111)

We can find the energy of the system at any timestep by adding the total work done onthe system in that time to the energy of the system at the previous step:

En+1 = En +Wtot (112)

Substituting (111) into (112) gives:

(PV )n+1 = (PV )n +Wcomp −∑

Wcyl (113)

Since the volume of the system remains constant, we can divide by V to get:

Pn+1 = Pn +Wcomp −

∑Wcyl

V(50)

Basic thermodynamics says that the work done by compression/expansion of gas atconstant pressure can be represented by W = P ·∆V . In each actuation of a cylindergas is compressed twice, once for extension and once for retraction. So for each timestep,the work done on the cylinder can be expressed as:

Wcyl = ((PAL)push + (PAL)pull) ·m (114)

L is constant, so it can be factored out. We can also substitute A for the effective pistonareas, giving:

Wcyl =

(πD2

4Ppush +

π(D2 − d2)

4Ppull

)L ·m (52)

Similarly, we can represent the work done by the compressor as:

Wcomp = Vcomp(P ) · dt · 1atm (51)

Note: since Vcomp is measured at atmospheric pressure, P must be 1 atm and not thesystem pressure

20

F Equivalent Radius/Load Derivation

To find the equivalent radius, we equate the rotational → linear motion transformationequations for both a wheel (15) and lead screw (61)

2πr · ω = v = ω · np (115)

We can then solve for r = req to find the radius that will provide the same rotational→ linear motion transformation as the given lead screw

req =np

2π(66)

We want to conserve the work done by a lead screw and equivalent wheel across onerotation (where x can be L or R).

W = 2πreq · Lx,eq = 2πTx (116)

Solving for Lx,eq, we get:

Lx,eq =Txreq

(117)

21