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Constant Force vs Ideal Spring to Return a Squiggle Motor's

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Constant Force vs Ideal Spring to Return a Squiggle Motor's Carriage
Summary
Using only steady state considerations with no air resistance, two systems of storing energy are
compared in combination with a Squiggle Motor, namely the SQL-RV-1.8. Gravity is used to represent a
constant force energy storage and is compared to an ideal Newtonian spring. It is shown that for a
practical case with ball bearings, the constant force method wastes less energy. Even though it loses
more to friction in the storage system, the gravity solution in providing a constant force, wastes less
energy to friction within the Squiggle Motor's self-locking screw, in another part of the system.
However since its losses are proportional to the force it must output, the constant force storage system
wastes more energy when a larger force is needed as with a slide bearing.
The System
Figure 1: Squiggle Motor & System
A Squiggle Motor and system by New Scale Technologies is shown in Figure 1. In the forward direction,
the squiggle motor must overcome friction as well as force energy into a storage component, which
powers the motion in the reverse direction.
Forward
πΉπ‘šπ‘œπ‘‘π‘œπ‘Ÿπ‘“π‘€π‘‘ = πΉπ‘“π‘Ÿπ‘–π‘π‘‘π‘–π‘œπ‘› + πΉπ‘ π‘‘π‘œπ‘Ÿπ‘Žπ‘”π‘’
In this analysis, let there be no air resistance, so this friction can be considered Newtonian and constant,
equal to the weight of its carriage times a coefficient of friction.
πΉπ‘“π‘Ÿπ‘–π‘π‘‘π‘–π‘œπ‘› = π‘š ∗ 𝑔 ∗ πœ‡1
Note that the system in Figure 1 utilizes compression springs to return the carriage pushed by the
motor. A coil spring of this type applies a force proportional to the distance it is compressed. To apply a
more constant force, consider using a mass on a cable. These two system components are shown in
Figure 2 where a red arrow indicates the direction in which a force is applied in order to store energy in
these components.
Figure 2: Components to Be Compared
For gravity, the force to store energy is the weight lifted plus some friction in the pulley element,
proportional to the tension in the cable, which for lifting at a constant speed will be the average of the
weight and force to store.
πΉπ‘ π‘‘π‘œπ‘Ÿπ‘Žπ‘”π‘’ = π‘Šπ‘’π‘–π‘”β„Žπ‘‘ + (πΉπ‘ π‘‘π‘œπ‘Ÿπ‘Žπ‘”π‘’ + π‘Šπ‘’π‘–π‘”β„Žπ‘‘) ∗
πΉπ‘ π‘‘π‘œπ‘Ÿπ‘Žπ‘”π‘’
πœ‡π‘”
2
πœ‡π‘”
π‘Šπ‘’π‘–π‘”β„Žπ‘‘ (1 + 2 )
=
πœ‡π‘”
(1 − 2 )
For the spring, the force to store energy is dependent on the current position of the carriage, x. The
zero of this x axis is the edge of the carriage's movement when there can be no less energy stored in the
spring. The spring constant k is the spring rate.
πΉπ‘ π‘‘π‘œπ‘Ÿπ‘Žπ‘”π‘’ = π‘˜ ∗ π‘₯ + πΉπ‘π‘Ÿπ‘’π‘™π‘œπ‘Žπ‘‘
Reverse
In the reverse direction, when the storage element is powering the movement of the carriage, and the
screw is simply retreating and controlling the motion, the motor and screw resist the motion using its
self-locking properties.
Figure 3: A Screw Analogy
Since it is self-locking, when an axial force is applied, it will hinder rotation, regardless of the direction.
For example, consider the self locking screw in Figure 3. It has a force F applied axially. A larger F
creates greater friction which causes the torque T required to unscrew to be larger as well, even though
the force F is attempting to push the screw to the right, the same direction as the unscrewing torque.
Note that due to the way the motor is used as a linear actuator, the friction with the screw in the
forward direction is already accounted for in the motor curves shown below in Figure 4.
Figure 4: Motor Performance Curves
Since there are no curves for a reverse load, the equivalent forward force, Feq, that causes the same
load on the motor as the reverse force, F, is calculated. In the diagrams in Figure 5, the two scenarios
are considered. The red represents the motor's nut and the green the screw. The angle  is the
arctangent of the screw pitch divided by the pitch circumference. Fm is the force of the motor at the
pitch radius, the torque on the nut divided by the pitch radius, and is the same for both scenarios to
make them equivalent. Rm is a reaction force that keeps the nut from being forced out, and is not the
same in both cases.
Figure 5
The force applied laterally to the screw is split to its components, but since the concern is the load on
the nut, only these interaction forces are shown; the green wedges are not complete free body
diagrams. The red wedges representing the nut are complete free body diagrams. The frictional forces
are kinetic since the screw is moving downward. Here summing forces in the horizontal directions,
assuming no acceleration for steady state, and equating the two Fm produces that
πΉπ‘’π‘ž = 𝐹
πœ‡ cos(πœƒ) − sin(πœƒ) + πœ‡2 cos(πœƒ) + πœ‡πœ‡2 sin⁑(πœƒ)
πœ‡ cos(πœƒ) + sin(πœƒ) + πœ‡2 cos(πœƒ) − πœ‡πœ‡2 sin(πœƒ)
If =2 this simplifies to
πΉπ‘’π‘ž = 𝐹
2πœ‡2 cos(πœƒ) − (1 − πœ‡22 )sin⁑(πœƒ)
2πœ‡2 cos(πœƒ) + (1 − πœ‡22 ) sin(πœƒ)
Having arrived at the conversion, which turns the force on the screw into a load against the motor,
πΉπ‘šπ‘œπ‘‘π‘œπ‘Ÿπ‘Ÿπ‘’π‘£
2πœ‡2 cos(πœƒ) − (1 − πœ‡22 )sin⁑(πœƒ)
=
(𝐹
− πΉπ‘“π‘Ÿπ‘–π‘π‘‘π‘–π‘œπ‘› )
2πœ‡2 cos(πœƒ) + (1 − πœ‡22 ) sin(πœƒ) π‘Ÿπ‘’π‘‘π‘’π‘Ÿπ‘›
For gravity, the force returned is the weight lifted minus the friction in the pulley element, proportional
to the tension in the cable, which for returning at a constant speed will be average of the weight and the
friction of the carriage.
πΉπ‘Ÿπ‘’π‘‘π‘’π‘Ÿπ‘› = π‘Šπ‘’π‘–π‘”β„Žπ‘‘ − (π‘Šπ‘’π‘–π‘”β„Žπ‘‘ + πΉπ‘“π‘Ÿπ‘–π‘π‘‘π‘–π‘œπ‘› ) ∗
πœ‡π‘”
πœ‡π‘”
πœ‡π‘”
= β‘π‘Šπ‘’π‘–π‘”β„Žπ‘‘ (1 − ) − πΉπ‘“π‘Ÿπ‘–π‘π‘‘π‘–π‘œπ‘›
2
2
2
For the spring, the force returned is the same as the force to store.
πΉπ‘Ÿπ‘’π‘‘π‘’π‘Ÿπ‘› = πΉπ‘ π‘‘π‘œπ‘Ÿπ‘Žπ‘”π‘’ = π‘˜ ∗ π‘₯ + πΉπ‘π‘Ÿπ‘’π‘™π‘œπ‘Žπ‘‘
Overall
The overall equations for load on the motor for the spring are thus:
Spring
πΉπ‘šπ‘œπ‘‘π‘œπ‘Ÿπ‘“π‘€π‘‘ = π‘šπ‘”πœ‡1 + π‘˜π‘₯ + πΉπ‘π‘Ÿπ‘’π‘™π‘œπ‘Žπ‘‘
πΉπ‘šπ‘œπ‘‘π‘œπ‘Ÿπ‘Ÿπ‘’π‘£
2πœ‡2 cos(πœƒ) − (1 − πœ‡22 ) sin(πœƒ)
=
(π‘˜π‘₯ + πΉπ‘π‘Ÿπ‘’π‘™π‘œπ‘Žπ‘‘ − π‘šπ‘”πœ‡1 )
2πœ‡2 cos(πœƒ) + (1 − πœ‡22 ) sin(πœƒ)
The overall equations for load on the motor for gravity are thus:
Gravity
πΉπ‘šπ‘œπ‘‘π‘œπ‘Ÿπ‘“π‘€π‘‘
πΉπ‘šπ‘œπ‘‘π‘œπ‘Ÿπ‘Ÿπ‘’π‘£
πœ‡π‘”
π‘Šπ‘’π‘–π‘”β„Žπ‘‘ (1 + 2 )
= π‘šπ‘”πœ‡1 +
πœ‡π‘”
(1 − 2 )
πœ‡π‘”
πœ‡π‘”
2πœ‡2 cos(πœƒ) − (1 − πœ‡22 ) sin(πœƒ)
=
(1
−
)
−
π‘šπ‘”πœ‡
(1
+
))
(π‘Šπ‘’π‘–π‘”β„Žπ‘‘
1
2
2
2πœ‡2 cos(πœƒ) + (1 − πœ‡22 ) sin(πœƒ)
Having the loads to look up the speeds on Figure 4 then converts this to motion. The speed is the rate of
change of x relative to time, and can be integrated numerically to obtain x.
Parameters
Forced Minimums
For a given situation, there is a minimum Weight or Fpreload required since the energy storage elements
must be able to move the carriage to that x=0 position where no less energy can be stored in the
element. For this, the Freturn must overcome the static friction of the carriage. If the static friction is
approximated as the kinetic friction times a constant C, Ffriction:
πΆπΉπ‘“π‘Ÿπ‘–π‘π‘‘π‘–π‘œπ‘› = π‘˜ ∗ 0 + πΉπ‘π‘Ÿπ‘’π‘™π‘œπ‘Žπ‘‘ = πΉπ‘π‘Ÿπ‘’π‘™π‘œπ‘Žπ‘‘
π‘ƒπ‘Ÿπ‘’π‘™π‘œπ‘Žπ‘‘π‘šπ‘–π‘› = πΆπΉπ‘“π‘Ÿπ‘–π‘π‘‘π‘–π‘œπ‘› = πΆπ‘šπ‘”πœ‡1
πΆπΉπ‘“π‘Ÿπ‘–π‘π‘‘π‘–π‘œπ‘›
πœ‡π‘”
1− 2
πœ‡π‘”
πœ‡π‘”
= π‘Šπ‘’π‘–π‘”β„Žπ‘‘ (1 − ) − πΆπΉπ‘“π‘Ÿπ‘–π‘π‘‘π‘–π‘œπ‘›
= π‘Šπ‘’π‘–π‘”β„Žπ‘‘
πœ‡π‘”
2
2
1+ 2
π‘Šπ‘’π‘–π‘”β„Žπ‘‘π‘šπ‘–π‘›
πœ‡π‘”
πœ‡π‘”
1+
2
2
= πΆπΉπ‘“π‘Ÿπ‘–π‘π‘‘π‘–π‘œπ‘›
πœ‡π‘” = πΆπ‘šπ‘”πœ‡1
πœ‡π‘”
1− 2
1− 2
1+
Other Parameters for Testing
The parameters vary based on the components used below are some parameters that will be used
along with the reasons for them.
g=9.81 m/s2 Gravity close enough to earth's surface
.16
πœƒ = atan (.529∗2πœ‹) = 2.75595° New Scale CAD Model1
m=.065 kg (Mass obtained from 3D modeling of items that need be moved)
πœ‡π‘” = .3
plastic-metal2
πœ‡1 as stated below in Table 1.2
πœ‡2 = 0.8 Steel-Steel3 with this kinetic approximated as static from source
k=.01 N/mm=10N/m closest available4 to .5gram/mm=.495N/m New Scale Recommended
Fpreload = Preloadmin as stated in Forced Minimums with C as stated below in Table 1.
Weight=Weightmin as stated in Forced Minimums with C as stated below in Table 1.
Table 1: Ball and Slide Bearing Parameters
πœ‡1
C
1
Ball Bearings
.002
2
Slide Bearings
0.2
1.2
http://www.newscaletech.com/downloads_registered/SQL-RV-1.8-6-12.STEP
http://www.tribology-abc.com/abc/cof.htm
3
http://www.tribology-abc.com/abc/screwlub.htm
4
http://www.leespring.com/compression_spec.asp?springType=C&forWhat=Search
2
Results
Simple linear/Euler approximated simulations are performed with constant time step of dt=.0001 sec,
the stated parameters and the typical load to speed relation in Figure 4, running the system from x=0 to
x=.006m and back.
Figure 6: Ball Bearing Case
Figure 7: Slide Bearing Case
Table 2: Work done by Motor
Ball Bearing
Slide Bearing
Gravity
0.00004428 J
0.00262 J
Spring
0.00038659 J
0.00219 J
The orders of magnitude of the work done by the motor are clearly different, reflecting the difference in
orders of magnitude of the coefficients of friction.
Conclusions
Since the losses induced by the additional friction are directly dependent on the force required, gravity
out performs the spring in the situation where a the force required is small and any additional force
provided is wasted. For the Squiggle motor, or on any other system where a self-locking screw is
interfaced similarly, the additional returning force of the spring a detriment, putting load on the motor.
In the cases that require a larger force however, a larger weight becomes the more detrimental, creating
more friction against the cable than a spring would have in the self-locking screw.
Appendix
An Additional Case
Unconventional Spring:
Assume it operates as an ideal spring, only with a custom k value equal to 0.5g/mm as recommended by
New Scale Technologies:
Figure 8: With Slide Bearings
Figure 9: With Ball Bearings
Speed Curve Fit used
3.3V typ case curvefit
70
60
50
40
30
20
10
y = 0.01162880x3 - 0.34213897x2 - 0.69805841x + 59.43689955
R² = 0.99996546
0
0
2
4
6
8
10
12
14
Figure 10
MATLAB code used
Speed.m:
function [ x ] = Speed(F)
%disp(F);
if(F<3) x=18; return; end
if(F>55) x=0; return; end
pol=[0.01162880, -0.34213897, -0.69805841, 59.43689955];
pol(4)=pol(4)-F;
q=roots(pol);
q2=q(q<=20);
x=q2(q2>=0);
if(length(x)>1) disp(x); x=x(1); end
end
gravvspring.m with parameters highlighted:
clear; clc; format compact; close all;
dt=.0001; %(sec) Euler Integration timestep
TotalCarriageMass=0.065;%m mass of carriage (kg)
Carriagefric=0.002; %mu-1 coefficient of friction on carriage
CarFricF=TotalCarriageMass*9.81*Carriagefric;
xlim=6/1000; %meters
C=2; %Static to kinetic friction ratio for slides
%Conversions for performance curve use
gramTOnewton=@(x) x*9.81/1000;
newtonTOgram=@(x) x*1000/9.81;
16
18
mmpsTOmps=@(x) x/1000;
mpsTOmmps=@(x) x*1000;
%Typ squiggle motor @3.3V Performance Curves
ForceFit=[0.01162880, -0.34213897, -0.69805841, 59.43689955];
Force=@(x) polyval(ForceFit,x);
%screw
ScrewFric=0.8; %mu-2
ScrewPit=.16/1000; %pitch m
ScrewPitR=.529/1000; %m
ScrewAngl=atan(ScrewPit/(2*pi*ScrewPitR)); %rad
ConversionRat=(2*ScrewFric*cos(ScrewAngl)-(1ScrewFric*ScrewFric)*sin(ScrewAngl))/(2*ScrewFric*cos(ScrewAngl)+(1ScrewFric*ScrewFric)*sin(ScrewAngl));
Fequiv=@(F)F*ConversionRat;
%gravity model
Gravpullfric=0.30; %mu-g
Weight=C*CarFricF*(1+Gravpullfric/2)/(1-Gravpullfric/2); %Min weight Newtons
Loadgfwd=@(x)CarFricF+Weight*(1+Gravpullfric/2)/(1-Gravpullfric/2); %Load fwd
in newtons
Loadgrev=@(x)Fequiv(Weight*(1-Gravpullfric/2)-CarFricF*(1+Gravpullfric/2));
%Load rev in newtons
%springd model
springk=10;%0.5*9.81; %0.5g/mm recommended by newscale 10N/m min from
leespring
preload0=C*CarFricF; %Min Fpreload
Loadsfwd=@(x)CarFricF+springk*x+preload0; %Load fwd in newtons
Loadsrev=@(x)Fequiv(springk*x+preload0-CarFricF); %Load rev in newtons
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% Gravcalcs
xg=[0];
tg=[0];
i=1;
Eg=0;
while(xg(i)<xlim)
Lgg(i)=newtonTOgram(Loadgfwd(xg(i)));
Lg(i)=newtonTOgram(Loadgfwd(xg(i)));
if(Lg(i)>65)fprintf('STALL%n'); end
Sg(i)=Speed(Lg(i));
Lg(i)=gramTOnewton(Lg(i));
SMg(i)=Sg(i);
Sg(i)=mmpsTOmps(Sg(i));
Eg=Eg+Lg(i)*Sg(i)*dt;
xg(i+1)=xg(i)+Sg(i)*dt;
tg(i+1)=tg(i)+dt;
i=i+1;
end
while(xg(i)<2*xlim)
Lgg(i)=newtonTOgram(Loadgrev(2*xlim-xg(i)));
Lg(i)=newtonTOgram(Loadgrev(2*xlim-xg(i)));
%if(Lg(i)>65)fprintf('STALL%n'); end
Sg(i)=Speed(Lg(i));
Lg(i)=gramTOnewton(Lg(i));
SMg(i)=Sg(i);
Sg(i)=mmpsTOmps(Sg(i));
Eg=Eg+Lg(i)*Sg(i)*dt;
xg(i+1)=xg(i)+Sg(i)*dt;
tg(i+1)=tg(i)+dt;
i=i+1;
end
Lgg(i)=newtonTOgram(Loadgrev(xg(i)));
Lg(i)=newtonTOgram(Loadgrev(xg(i)));
Sg(i)=Speed(Lg(i));
SMg(i)=Sg(i);
Lg(i)=gramTOnewton(Lg(i));
Sg(i)=mmpsTOmps(Sg(i));
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% springcalcs
xs=[0];
ts=[0];
i=1;
Es=0;
while(xs(i)<xlim)
Lsg(i)=newtonTOgram(Loadsfwd(xs(i)));
Ls(i)=newtonTOgram(Loadsfwd(xs(i)));
if(Ls(i)>65)fprintf('STALL%n'); break; end
Ss(i)=Speed(Ls(i));
Ls(i)=gramTOnewton(Ls(i));
SMs(i)=Ss(i);
Ss(i)=mmpsTOmps(Ss(i));
Es=Es+Ls(i)*Ss(i)*dt;
xs(i+1)=xs(i)+Ss(i)*dt;
ts(i+1)=ts(i)+dt;
i=i+1;
end
while(xs(i)<2*xlim)
Lsg(i)=newtonTOgram(Loadsrev(2*xlim-xs(i)));
Ls(i)=newtonTOgram(Loadsrev(2*xlim-xs(i)));
if(Ls(i)>65)fprintf('STALL%n'); break; end
Ss(i)=Speed(Ls(i));
Ls(i)=gramTOnewton(Ls(i));
SMs(i)=Ss(i);
Ss(i)=mmpsTOmps(Ss(i));
Es=Es+Ls(i)*Ss(i)*dt;
xs(i+1)=xs(i)+Ss(i)*dt;
ts(i+1)=ts(i)+dt;
i=i+1;
end
Ls(i)=newtonTOgram(Loadsrev(2*xlim-xs(i)));
Ss(i)=Speed(Ls(i));
Ls(i)=gramTOnewton(Ls(i));
SMs(i)=Ss(i);
Ss(i)=mmpsTOmps(Ss(i));
plot(xg,Lg,xs,Ls); xlabel('dist moved from start (m)');
ylabel('load opposing motor (N)'); legend('Gravity','Spring');
fprintf('Work done by motor against friction and energy storing load in
system:\nGravity %.8f J\nSpring %.8f J\n',Eg,Es);
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