Design of recumbent elliptical trainer

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DESIGN OF A RECUMBENT ELIPTICAL TRAINER.

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1. RESULTS AND ANALYSIS
TASK 1
A recumbent elliptical machine is used to rehabilitate people with restricted mobility in their
lower extremities. The recumbent elliptical machine trajectory is supposed to be elliptical and
this can only be achieved by the use of linkages, CAMs or a combination of both. In this
particular task we have used a 4 bar linkage mechanism to get the desired elliptical trajectory.
A 4 bar linkage mechanism is known to magnify forces, convert rotational motion to linear
motion (in our case, we can manipulate the linear motion to produce an elliptical trajectory), and
also a 4 bar linkage can be used to constrain motion, e.g. in the knee joint. The velocity and
accelerations are obtained by consideration of the velocity of points on a link which moves in a
perpendicular direction to the link.
So as to come up with an efficient recumbent machine, it is necessary for us to consider mobility
(degrees of freedom). The degrees of freedom can be defined as the modes in which a device can
move. Mobility in this case can be equated to the total aspects of motion. The Gruebler’s
equation can be used to get the mobility of a mechanism. In this case, the 4 bar linkage
mechanism used had a mobility of 1.
Mobility
GRUEBLER’S EQUATION

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𝑓 = 3(𝑛 − 1) − 2𝑛
Where;
n is the total number of links
f1 is the total number of joints
Since it is a 4 bar linkage
n = 4, f1 = 4
𝑓 = 3(4 − 1) − 2(4)
F=9-8
F=1
Hence, the resulting mobility or degree of freedom is 1.
Hence, the resulting mobility or degree of freedom is 1.
TASK 2

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In order to come with a suitable elliptical recumbent machine that operates under the 4 bar
mechanism, then there is a need to understand how we procedurally come up with the final
design drawings to be used. First, it is quintessential for one to note that a 4 bar linkage is made
up of four bars and they are, the coupler, the crank, the follower, and the ground.
It is also important for one to note that the place where one is supposed to place the legs
(peddles), during the back-and –forth motion of a 4-bar linkage mechanism, will trace an
elliptical motion.
The elliptical motion is what we actually desire to achieve in an elliptical recumbent machine for
people with trouble in their lower extremities. Below is the procedure followed during the design
of the elliptical recumbent machine, and using the engineering drawing knowledge we were able
to trace the elliptical trajectory whilst catering for the leg space.
DESIGN PROCEDURE
a. Begin by drawing a coupler in both its initial and final positions
b. Then after establishing the ground and drawing the coupler, go ahead and draw arcs from
the mounting points of the output bar and ensure that the radius are the same for each
mounting hole.
c. Then draw a pair of locus lines connecting the intersections of arcs.
d. Finally, draw the connecting bars.
Ergonomics, effectiveness and efficiency are the three crucial factors that determine the
mechanism dimensions. Also checking on the existing elliptical recumbent design one can be
able to narrow down at least to suitable dimensions.
Kinematic synthesis,
𝑏2
= 𝑎2
+ 𝑐2
− 2𝑎𝑐𝑠𝑖𝑛𝜃 − 𝑠2
− 2𝑎𝑠𝑐𝑜𝑠𝜃

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0.82
− 0.22
− 0.172
− 2(0.2 ∗ 0.17)𝑠𝑖𝑛64° = −𝑠2
− 2(0.2)𝑠 ∗ 𝑐𝑜𝑠64
𝑠2
+ 0.17525 − 0.501 = 0
(−0.1753(+ 𝑜𝑟 −)√[0.17532 − 4(−0.5010)])/2 = 0.6256 𝑚

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TASK 3
In order to analyse our solution for this particular assignment it is quite crucial that we get both
the acceleration and the velocity for our 4 linkage bar mechanism. The velocity and acceleration
can be derived using the velocity and acceleration diagrams. Below we have in depth tried to
explain how we derived both the two solutions using the velocity and acceleration diagrams.
Velocity diagram procedure
a. Calculation of the tangential velocity is the first step while coming up with a velocity
diagram.

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b. Then draw a vector from point O to A.
c. Next we have a vector ab since velocity of B is relative to A.
d. Velocity of B is absolute to O and hence the vector OB starts at a.
Acceleration diagrams
The relationships for acceleration are similar to those of velocity; acceleration diagram is
obtained from the velocity diagram. When drawing the acceleration diagram, one needs to
consider the two components of acceleration:
a) Radial acceleration-Centripetal acceleration
b) Tangential acceleration
Drawing the Acceleration diagram
Acceleration of any point say for example O on the link AB will be obtained by dropping a
perpendicular line from the link AB on the velocity diagram.
i. Calculate all the possible acceleration before embarking on the construction of the
acceleration diagram.
a) The tangential acceleration of B relative to A
b) The centripetal acceleration of B relative to A, 𝛼 = 𝜔2AB
c) The tangential acceleration of C relative to B is unknown and will be measured
from the diagram after construction
ii. First draw the centripetal acceleration of link AB, obtained from b) 𝛼 = 𝜔2AB,
above.
iii. Add the centripetal acceleration of link BC, because there are two acceleration for
point C, name one c1

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iv. Add the tangential acceleration of point C relative to B, designate C1C, the direction
of this link is to right angles of the previously drawn vector, we call this new line C
line-dotted line.
v. Draw the acceleration of the slider, note that it is constrained to the horizontal
direction, therefore the vector starts at and must intersect the C line, name this point
of intersection C.
vi. Obtain by measurement the acceleration of C relative to B and calculate the
centripetal and tangential accelerations.

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From the velocity diagram,
𝑣(𝑏) = 𝑤𝐴𝐵

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5.24 ∗ 0.2 =
1.048𝑚
𝑠
Vc (measured from the velocity diagram) = 8.3m/s
From the acceleration diagram
∝ (𝑏) = 𝑤2
𝐴𝐵
5.242
∗ 0.2 =
5.49𝑚
𝑠
∝ (𝑐) = 𝑎𝑙𝑠𝑜 𝑚𝑒𝑎𝑠𝑢𝑟𝑒𝑑 𝑓𝑟𝑜𝑚 𝑡ℎ𝑒 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝑑𝑖𝑎𝑔𝑟𝑎𝑚
Centripetal = 8.0m/s2
Tangential = 8.01m/s2
TASK 4
Mechanical Advantage
This is the ratio of output force to the input force of a linkage. In an ideal linkage the speed ratio
and the mechanical advantage are defined so they yield the same number. In obtaining the
mechanical advantage of the four-bar linkage, we calculate basing on the principle of virtual work,
where power in is equal to power out. Mechanical advantage is usually of importance to a design.
It gives the relationship between force and position. Force or power or torque requirement can be
converted to a velocity ratio, the velocity analysis can then be used in the design of a mechanism
required to obtain the mechanical advantage. The velocity ratio is a function of the position of a
mechanism and hence in extension the mechanical advantage is a function of mechanical position.
The formula is used to calculate the mechanical advantage y considering the movements and
positions of the various cranks and angles. We input the values for the angles and the lengths of
the cranks, coupler and the ground link in the formula to obtain the mechanical advantage.

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The Mechanical advantage of a four-bar linkage mechanism is equal to its speed ratio. The
principle of virtual work is applied in the determination of the mechanical advantage of a four-bar
linkage system where, the Power input equals the power output (Toussaint, 2003).
The figures below will try and illustrate this analysis.
Fig-3 Inertia force, i) a translating body ii) Compound pendulum, with an angular acceleration iii)
inertia force and couple acting on the pendulum
The mechanical advantage of a 4R linkage equals its speed ratio

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From the principle of virtual work 𝑇(𝑖𝑛)𝜃 = 𝑇(𝑜𝑢𝑡)𝜑
𝑅 =
𝜃
𝜑
=
𝑇(𝑜𝑢𝑡)
𝑇(𝑖𝑛)
=
−𝑎𝑏𝑐𝑜𝑠𝜓𝑠𝑖𝑛Ө − 𝑏𝑔𝑠𝑖𝑛𝜓 + 𝑎𝑏𝑐𝑜𝑠Ө𝑠𝑖𝑛𝜓
𝑎𝑔𝑠𝑖𝑛Ө + 𝑎𝑏𝑐𝑜𝑠𝜓𝑠𝑖𝑛Ө − 𝑎𝑏𝑐𝑜𝑠Ө𝑠𝑖𝑛𝜓
−0.2 ∗ 0.16𝑐𝑜𝑠90𝑠𝑖𝑛81 − 0.16 ∗ 0.83𝑠𝑖𝑛90 + 0.2 ∗ 0.16𝑐𝑜𝑠81𝑠𝑖𝑛90
0.2 ∗ 0.83𝑠𝑖𝑛81 + 0.2 ∗ 0.16𝑐𝑜𝑠90𝑠𝑖𝑛81 − 0.2 ∗ 0.16𝑐𝑜𝑠81𝑠𝑖𝑛90
= −
0.127794097
0.158950361
= −0.80398746
Mechanical advantage
2nd
class lever >1
3rd
<1
1st
>1 or <1
TASK 5
Dynamic force in science can be described as the amount of speed or acceleration required to set
an object into motion. In simpler terms, it is the process through which the energy necessary
during the moment of moving an object is either ameliorated or reduced.
This particular task is supposed to enable us break down and understand those factors that
influence the increase or reduction of motion in an elliptical recumbent machine. From this

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particular factors we have calculated the speed and acceleration needed to set our machine into
motion. This has been done using the D’Alembert principle.
D’Alembert principle can be used to reduce a dynamic system into an equivalent static system
and utilize the equations and techniques in static force analysis in studying the system. A slider
crank method as is our case, converts rotary motion to straight line motion. According to
D’Alembert Principle, a body can be brought into an equilibrium position by the application of a
force that is equal to F = ma and in a direction that is opposite to the direction of acceleration.
Dynamic analysis of a slider crank mechanism
Position of the slider at angle 81˚,
𝑥 = 𝑟(1 − 𝑐𝑜𝑠𝜃)
𝑋 = 0.2(1 − cos 81°)
= 0.1687 m
Velocity of the slider wheel,
𝑣 = 𝑟𝑤𝑠𝑖𝑛𝜃
𝑣 = 0.2 ∗ 2𝜋 ∗ (
81
60
) 𝑠𝑖𝑛81°
= 1.6756 m/s
Acceleration of slider wheel,
∝= 𝑟𝑤2
(1 −
1
𝑛
)
∝= 0.2 (2 ∗ 3.142 ∗
81
60
)
2
(1 −
1
0.8
)
= -3.5975 m/s2

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TASK 6
A gear train is a mechanical system that is composed of gears mounted on a frame and engaged by
the teeth of the gears. Gear trains have the ratio of their pitch circles defining the speed ratio and
the mechanical advantage of the gear set and a planetary gear system used to provide a high gear
reduction. To achieve speed reduction using gears, a smaller gear/ pinion gear is connected to a
large gear or final drive gear. Shafts are attached to these gear drives and at the end, linkages
connected to complete the mechanism of transfer of motion and the reduction system for that
matter (Hartenberg & Denavit, 1964).
A gear drive is usually embraced when the distance between the pinion and the final drive is
quite small. Gears are always preferred in power transmission since they transmit exact velocity
ratio. Gears have high efficiency when compared to other power transmission, and it can be used
in transmission of large power. Other advantages of gear drives include; reliable services and
compact layouts.
The gear drive that we tackled is made up of two gears (the pinion and the final drive), this was
so because we had to be economical and yet effective with the design. The pinion will be directly
connected to the pinion gear using a shaft, and the rotational motion from the pinion gear will be
transmitted to the final drive gear that serves as a step down, thus reducing the rotational speed.
The final drive gear has 2 shafts attached on it and when they receive the motion they take it to a
twice-the-diameter of the gear so as to slash by half the 100rpm received from the final gear.
Each of the two plates has a linkage attached to it and they are rotated at a speed of 50rpm.

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Pressure angle – it can be explained as the angle between the normal to 2 gear teeth at that point
where they are in contact with one another.
Addendum – referred to as the distance of a gear tooth from the pitch circle to the uppermost
part of the tooth.
Dedendum – described as the radial distance of a gear tooth straight from the bottom of the
tooth to the pitch center.

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Gear drive
From catalogue
Pressure angle = 20º
Pitch radius = pinion, 10 and gear, 50
Outside diameter = pinion, 22 and gear, 102
Pinion gear no. of teeth = 20
Final gear no. of teeth = 100
𝑐𝑖𝑟𝑐𝑢𝑙𝑎𝑟 𝑝𝑖𝑡𝑐ℎ =
𝜋 ∗ 100
100
=
𝜋 ∗ 20
20
= 1.0
Diametric pitch = 𝑝(𝑑) =
𝑇
𝐷
=
𝜋
𝑝(𝑐)
Pinion =
𝜋
1.0
= 3.1416
Follower =
𝜋
1.0
= 3.1416
Module =
𝑝(𝑐)
𝜋
=
𝐷
𝑇
Pinion module =
20
20
= 1

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Gear module =
100
100
= 1
ADENDUM
Pinion, T = 20; Ø = 20º m=1mm
P(c) = 𝜋𝑚
Length of arc of contact =?
Arc of contact (assume) = 1.75 times the circular pitch = 1.75 ∗ 3.1416 = 5.4978𝑚𝑚
Length of path of contact =𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑎𝑟𝑐 𝑜𝑓 𝑐𝑜𝑛𝑡𝑎𝑐𝑡 ∗ 𝑐𝑜𝑠∅ = 5.4978𝑐𝑜𝑠20° = 5.166𝑚𝑚
Let RA = rA = radius of the addendum circle of each wheel
We know that pitch circle radii of each wheel;
𝑅 = 𝑟 =
𝑚 ∗ 𝑇
2
=
1 ∗ 20
2
= 10𝑚𝑚
5.166 = √ [(𝑅(𝐴))2
– 𝑅2
cos2
Ø] + √ [(𝑟𝐴)2
− 𝑟2
cos2
Ø − (𝑅 + 𝑟)]
5.166
2
= √[(𝑅)2 − (10)2 cos2 20˚] − 10𝑠𝑖𝑛20˚
(2.583 + 3.42)2
+ 76.6 = 𝑅(𝐴)2
𝑅 (𝐴) = 10.61𝑚𝑚
10.61 – 10 = 0.61𝑚𝑚
Gear
T=100, Ø=20, m=1mm
Pc = 𝜋 ∗ 1
Length of arc of contact =?
Taking the arc of contact as 1.75 we get,
1.75 ∗ 3.1416 = 5.4978𝑚𝑚
Length of path of contact = length of arc of contact*cos Ø

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5.4978 ∗ 𝑐𝑜𝑠 20 = 5.166𝑚𝑚
Let RA = Ra = radius of the addendum circle of each wheel
We know that pitch circle radii of each wheel,
𝑅 = 𝑟 = (𝑚 ∗ 𝑇)/2
1 ∗ 100/2 = 50𝑚𝑚
5.166 = √[(𝑅𝐴)2
− 𝑅2
𝑐𝑜𝑠2
Ø] + √[(𝑟𝐴)2
− 𝑟2
𝑐𝑜𝑠2
Ø − (𝑅 + 𝑟) − 𝑠𝑖𝑛Ø]
5.166
2
= √[(𝑅𝐴)2
− 502
𝐶𝑂𝑆2
20 − 50𝑠𝑖𝑛20]
2.583 = √[(𝑅𝐴)2
− 1915.11] − 17.101
𝑅𝐴2
= 2302.69
𝑅𝐴 = 47.99𝑚𝑚
𝑅 − 𝑅𝐴 = 2.0145𝑚𝑚
DEDENDUM
Dedendum = pitch circle diameter * cos Ø
20𝑐𝑜𝑠20˚ = 18.794
Pinion Dedendum =20 – 18.794 = 1.206 𝑚𝑚
Gear root circle diameter = 100𝑐𝑜𝑠20˚ = 93.94
Gear Dedendum = 100 – 93.94 = 6.06 𝑚𝑚
Base radius (pinion) = 18.794/2 = 9.397 𝑚𝑚
Base radius (gear) = 93.97/2 = 46.99 𝑚𝑚
TASK 7

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REFERENCES
Toussaint, G. T., "Simple proofs of a geometric property of four-bar linkages," American
Mathematical Monthly, June/July 2003, pp. 482–494.
Myszka, David (2012). “Machines and Mechanisms: Applied Kinematic Analysis. New Jersey:
Pearson Education.” ISBN 978-0-13-215780-3
Chakrabarti, Amaresh (2002). Engineering Design Synthesis: Understanding, Approaches and
Tools. Great Britain: Springer-Verlag London Limited. ISBN 1852334924
Hartenberg, R.S. & J. Denavit (1964) “Kinematic synthesis of linkages, New York: McGraw-
Hill”, online link from Cornell University.

Design of recumbent elliptical trainer

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