Note: All catalog parts are shown in the "Cost & Partslist" tab
Sample Calculations 1
Engine Input
Maximum torque: 200 ftlbs
Maximum rpm: 6000 rpm
Note: Specs. for engines can be found here, link.
Cones
Driving Cone:
small dia. = 2.86 in (radius = 1.43 in)
large dia. = 7.40 in
angle of cone = 20 °
length of cone = [7.40  2.86 /[tan 20°]] + .625 (width of trans. belt) = [6.24 + .625] in = 6.86 in
Driven Cone:
small dia. = 6.44 in
large dia. = 10.98 in
angle of cone = 20 °
length of cone = [10.98  6.44 /[tan 20°]] + .625 (width of trans. belt) = [6.24 + .625] in = 6.86 in
Transmission ratio
Transmission ratio range available = 2.86 : 10.98 to 7.40 : 6.44 = 0.26 to 1.15 = 1 to 4.41
Teeth of cones & transmission belt
maximum torque = 200 ftlbs * [12 in/1 ft] = 2400 lbsin
maximum pulling force = 2400 lbsin / 1.43 in (small dia. of driving cone) = 1678.3 lbs
Transmission Belt
Below is some catalog data for a cog belt, it is from this website: www.gates.com
Sample Calculations 1
Engine Input
Maximum torque: 200 ftlbs
Maximum rpm: 6000 rpm
Note: Specs. for engines can be found here, link.
Cones
Driving Cone:
small dia. = 2.86 in (radius = 1.43 in)
large dia. = 7.40 in
angle of cone = 20 °
length of cone = [7.40  2.86 /[tan 20°]] + .625 (width of trans. belt) = [6.24 + .625] in = 6.86 in
Driven Cone:
small dia. = 6.44 in
large dia. = 10.98 in
angle of cone = 20 °
length of cone = [10.98  6.44 /[tan 20°]] + .625 (width of trans. belt) = [6.24 + .625] in = 6.86 in
Transmission ratio
Transmission ratio range available = 2.86 : 10.98 to 7.40 : 6.44 = 0.26 to 1.15 = 1 to 4.41
Teeth of cones & transmission belt
maximum torque = 200 ftlbs * [12 in/1 ft] = 2400 lbsin
maximum pulling force = 2400 lbsin / 1.43 in (small dia. of driving cone) = 1678.3 lbs
Transmission Belt
Below is some catalog data for a cog belt, it is from this website: www.gates.com
For a CVT 1 that uses cones with two opposite teeth, only one tooth is engaged with the transmission belt most of the time. If it is desired to have more teeth engaged, then a cone torque transmitting members, which we have patented, or a cone with one or two opposite toothed sections can be used.
For a cone with two opposite toothed sections, for each toothed section the teeth should be oriented so that at all axial sections of a cone the distance between the teeth is constant, and so that the distance between the teeth for all axial sections of a cone are identical. In order to satisfy this requirement, the surfaces of some teeth have to be curved. For example, if one tooth is straight, then the teeth adjacent to the straight tooth have to be curved in order to ensure that the distances between the straight tooth and its adjacent teeth remain constant for all axial sections, which have different diameters, of the cone.
Since here the curvature of the teeth is not constant, for a cone with two opposite toothed sections, a transmission belt that has round teeth (the teeth have round crosssections when viewed from the bottom of the belt) has to be used, see Figs. 1 & 2. Here full surface to surface engagement contact between a tooth of a cone and a tooth of a transmission belt cannot be achieved. Even if the teeth are made out of rubber, if you press a flexible round rubber cable on a table, you will see that the contact area is still very small (certainly less than 1/16 in).

A cone with two opposite toothed sections will also be more expensive than a cone with one tooth, and will allow a shorter duration for changing the axial position of a cone.
A transmission belt for a cone is shown as sideviews in Figs. 3 and 5, and as a sectional frontview in Fig. 4. It comprises of tooth plates, which are preferably made out of metal or a strong composite material, on which each a tooth is shaped; reinforcement; and rubber that is bonded to the tooth plates and the reinforcement.
If the pulling force for the transmission belt is 2000 lbs, then the transmission belt with the dimensions shown in Figs. 6 & 7 can be used. For a 2000 lbs pulling force, the minimum required engagement area is 2000 lbs / 30,000 psi = 0.067 in^2, if steel tooth plates are used. For the transmission belt of Figs. 6 & 7, the contact area of a tooth of the transmission belt is 0.1 in * 1.5 in = .15 in^2; and the shear resisting area of a tooth plate is also 0.1 in * 1.5 in = .15 in^2.
In catalog data for a Gates 8mm Pitch Poly Chain GT Carbon belt, it is listed that a 36mm (1.42 in) wide belt can transmit up to 216.7 HP. Therefore, we can assume that the rubber portion of the 1.5 in wide transmission belt shown in Figs. 6 & 7 is strong enough to transmit at least 200 HP.
A transmission belt for a cone is shown as sideviews in Figs. 3 and 5, and as a sectional frontview in Fig. 4. It comprises of tooth plates, which are preferably made out of metal or a strong composite material, on which each a tooth is shaped; reinforcement; and rubber that is bonded to the tooth plates and the reinforcement.
If the pulling force for the transmission belt is 2000 lbs, then the transmission belt with the dimensions shown in Figs. 6 & 7 can be used. For a 2000 lbs pulling force, the minimum required engagement area is 2000 lbs / 30,000 psi = 0.067 in^2, if steel tooth plates are used. For the transmission belt of Figs. 6 & 7, the contact area of a tooth of the transmission belt is 0.1 in * 1.5 in = .15 in^2; and the shear resisting area of a tooth plate is also 0.1 in * 1.5 in = .15 in^2.
In catalog data for a Gates 8mm Pitch Poly Chain GT Carbon belt, it is listed that a 36mm (1.42 in) wide belt can transmit up to 216.7 HP. Therefore, we can assume that the rubber portion of the 1.5 in wide transmission belt shown in Figs. 6 & 7 is strong enough to transmit at least 200 HP.
In Figs. 3 to 5, a half a cylinder tooth shape is used. It is based on the tooth shape of a roller chain. Other tooth shapes can also be used. If smooth engagement can be achieved, then it is desirable to use a slightly narrower tooth shape, see Fig. 8, than a half a cylinder tooth shape.
In order to help prevent a tooth of a transmission belt from disengaging with the tooth it is engaged with, a small intrusion can be shaped into each tooth of the transmission belt; an example of this method is shown in Fig. 9. During engagement, into the small intrusion of each tooth of the transmission belt that engages with a tooth of their cone, a small protrusion of the tooth of their cone is inserted; an example of a small protrusion of a tooth of a cone is shown in Fig. 10.
In order to allow for smooth engagement, it is recommended that the small intrusions are each shaped on the front surface of their tooth (as shown for the tooth shape shown in Fig. 9, which is moving in the right to left direction). Experimentations (trial and error, simulations, etc) and science can be used to determine “the allowable surface of a tooth (front surface, back surface, or both)” and “the ideal size and shape” for a small intrusion and a small protrusion.
Since the transmission belt shown in Figs. 3 to 5 is tapered, for the larger side surface of the transmission belt, the neutralaxis of the transmission belt is positioned at a farther distance relative to the teeth of the transmission belt so that a greater moment due to the force on a tooth is applied to the transmission belt. But here, this greater moment is compensated by a larger counteracting moment due to the pretension in the slack side of the transmission belt, so this should not be a problem.
If desired, as an option and not as a necessity, the larger side surface of the transmission belt shown in Figs. 3 to 5 can be diagonally reinforced. An example where wire reinforcement is used to diagonally reinforce the larger side surface of a transmission belt is shown as a sideview in Fig. 11, and as a sectional frontview in Fig. 12. Here the larger side surface facing side surface of each tooth plate has a lower anchor plate and an upper anchor plate, which both are used for attaching or threading through a reinforcement wire. When a reinforcement wire is threaded through the lower anchor plates and upper anchor plates, then it is recommended that the reinforcement wire is properly bonded to each lower anchor plate and to each upper anchor plate; in order to do this, high strength adhesives can be used.
In order to help prevent a tooth of a transmission belt from disengaging with the tooth it is engaged with, a small intrusion can be shaped into each tooth of the transmission belt; an example of this method is shown in Fig. 9. During engagement, into the small intrusion of each tooth of the transmission belt that engages with a tooth of their cone, a small protrusion of the tooth of their cone is inserted; an example of a small protrusion of a tooth of a cone is shown in Fig. 10.
In order to allow for smooth engagement, it is recommended that the small intrusions are each shaped on the front surface of their tooth (as shown for the tooth shape shown in Fig. 9, which is moving in the right to left direction). Experimentations (trial and error, simulations, etc) and science can be used to determine “the allowable surface of a tooth (front surface, back surface, or both)” and “the ideal size and shape” for a small intrusion and a small protrusion.
Since the transmission belt shown in Figs. 3 to 5 is tapered, for the larger side surface of the transmission belt, the neutralaxis of the transmission belt is positioned at a farther distance relative to the teeth of the transmission belt so that a greater moment due to the force on a tooth is applied to the transmission belt. But here, this greater moment is compensated by a larger counteracting moment due to the pretension in the slack side of the transmission belt, so this should not be a problem.
If desired, as an option and not as a necessity, the larger side surface of the transmission belt shown in Figs. 3 to 5 can be diagonally reinforced. An example where wire reinforcement is used to diagonally reinforce the larger side surface of a transmission belt is shown as a sideview in Fig. 11, and as a sectional frontview in Fig. 12. Here the larger side surface facing side surface of each tooth plate has a lower anchor plate and an upper anchor plate, which both are used for attaching or threading through a reinforcement wire. When a reinforcement wire is threaded through the lower anchor plates and upper anchor plates, then it is recommended that the reinforcement wire is properly bonded to each lower anchor plate and to each upper anchor plate; in order to do this, high strength adhesives can be used.
The specification for the transmission belt and the locations of the lower anchor plates and upper anchor plates should be selected such that the transmission belt can properly bent upwards and bent downwards as needed; and so that the reinforcement wire(s) are never slack when the transmission belt is bent downwards as required, so that the reinforcement wire(s) can provide some resistance to the force applied on a tooth. The reinforcement wire(s) do not have to be never slack when the transmission belt is bent upwards, if the teeth of the transmission belt are not used to transmit a large torque when the transmission belt is bent upwards. The proper specification for the transmission belt and the locations of the lower anchor plates and upper anchor plates to satisfy the criteria of this paragraph and any other requirements can be obtained through experimentations (trial and error, simulations, etc) and science.
If needed/desired tensioning pins can shaped/attached to the diagonally reinforced side surface of a transmission belt (see Figs. 11 & 12) directly or through the use of other means such as inserted metal pins for example. The tensioning pins shown in Figs. 11 & 12 can be used to help maintain tension (remove slack) in the reinforcement wire(s) as the transmission belt is bent downwards. Similarly tensioning pins that help maintain the tension in the reinforcement wire(s) as the transmission belt is bent upward can also be used. The proper specification for the transmission belt, the locations of the lower anchor plates and upper anchor plates, and the ideal size and positions of the tensioning pins, to satisfy the criteria of the previous paragraph and any other requirements can be obtained through experimentations (trial and error, simulations, etc) and science.
If desired, the smaller side surface or any side surface of a transmission belt, whether it is tapered or not, can also be diagonally reinforced using reinforcement wire(s) in the same manner as described in the previous paragraph. If both side surfaces of a transmission belt are diagonally reinforced, then it also needs to be ensured that the transmission belt can properly bent upwards and bent downwards as needed without twisting.
Axial movements required for trans. dia. change of 2 teeth
Tooth:
dia. of a tooth =.25 in
width of tooth = .375 in
Driving Cone:
circum. of small dia. = π * d = 3.14 * 2.86 = 8.98 in
number of teeth at small dia. = 8.98 in / .375 in = 24 teeth
circum. of large dia. = π * d = 3.14 * 7.44 = 23.25 in
number of teeth at large dia. = 23.25 in / .375 in = 62 teeth
for a distance of 6.24 in (length – width of belt), the number of teeth increase from 24 to 62
therefore, axial distance for 1 tooth = 6.24 in / [62 teeth – 24 teeth] = 0.16 in
axial distance for 2 teeth = 0.16 in * 2 = 0.33 in
number of teeth at large dia.  number of teeth at small dia. = 62 24 = 40 teeth
Driven Cone:
circum. of small dia. = π * d = 3.14 * 6.44 = 20.23 in
number of teeth at small dia. = 20.23 in / .375 in = 54 teeth
circum. of large dia. = π * d = 3.14 * 10.98 = 34.49 in
number of teeth at large dia. = 23.25 in / .375 in = 92 teeth
number of teeth at large dia.  number of teeth at small dia. = 92 54 = 40 teeth
Total transmission ratios available:
Total transmission ratios = [40 teeth (=62 – 24 or 92 54)]/”2” * “3” = 57 transmission ratios
Note: “2” since only an even number of teeth can be used for the cones, & “3” since for each axial position change 3 transmission ratios can be obtained, see “The CVT” tab.
Force needed to move a cone axially due to inertia
Below is the calculation where a force F is used to move a cone of mass m:
F = m * a, hence a = F/m,
where F  axial force on cone , m  mass of cone, a  axial acceleration of cone
Lets test an axial force of 500 lbs (although this force seems large, it is manageable, please bear with us) and assume a cone and its mover mechanism weights 2 kg, then:
F = 500 lbs = 2225 N
m = 2 kg
a = F/m = 2225 N / 2 kg = 1112.5 m/s^2
In order to determine the duration that a cone can be moved axially we use:
ɵ = ω * t, from which we get t = ɵ/ω
where:
ɵ  angular rotation where the fastest cone (driven cone) can be moved
ω – maximum rpm of the fastest cone
t  duration that the fastest cone can be moved
For a 6000 rpm input, and 7.4 in dia. driving cone & a 6.44 in dia. driven cone, we get:
ω = 6000 * 7.4/6.44 rpm = 6894.4 rpm = 6395.4/[60 rot/sec] = 114.9 rot/sec = 114.9 *360°/sec = 41366.5°/sec
If we allow 170° of rotation to move the fastest cone, then we get:
t = ɵ/ω = 170° / 41366.5°/sec = 0.00411 sec
In order to determine the movement of the fastest cone due to the axial force F, we use:
s = v * t + 1/2 * a * t^2
where:
s  axial distance traveled by cone
v  initial axial velocity of cone,
a  acceleration of cone due to force F
t  duration that the axial force F can be applied
Using the values we calculated earlier we get:
s = 0 * 0.00411 sec + 1/2 * 1112.5 m/sec^2 * [0.00411 sec]^2 = 0.0094 m = (0.0094 * 39.37) in = 0.37 in
Since the amount of axial movement required is 0.33 in, and since 0.37 in > 0.33 in this will work.
Force needed to move a cone axially due to tension and friction
Force needed due to tension of transmission belt:
tension (T)= 1678.3 lbs
max. axial force needed due to axial force of T = 2 * T * tan 20° = 2 * 1678.3 * 0.36 = 1221.7 lbs
(Note: if a Vbelt pulley, as in an iCVT, with an angle of 70° is used as then:
max. axial force needed due to axial force of T = [2 * T * tan 70°] * 1.8 = [2 * 1678.3 * 2.74] * 1.8 = 16600.1 lbs.
The multiplication of 1.8 is used because an iCVT can have almost two half transmission belt/chain loops under tension, while a CVT 1 only has a slightly greater than half transmission belt/chain loop under tension.)
Force needed due to friction of transmission belt:
max. normal force of T = 2 * T / cos 20° = 3572.1 lbs
(Note: if a Vbelt pulley, as in an iCVT, with an angle of 70° is used as then:
max. normal force of T = [2 * T / cos 70°] * 1.8 = 9814.2 lbs * 1.8 = 17665.6 lbs
The multiplication of 1.8 is used because an iCVT can have almost two half transmission belt/chain loops under tension, while a CVT 1 only has a slightly greater than half transmission belt/chain loop under tension.)
max. axial force needed due to friction = µ *max. normal force of T (due to belt resting on cone) + µ * T (due to tooth) = 0.1 * 3572.1 lbs + 0.1 * 1678.3 lbs = 525.0 lbs
Total force needed due to transmission belt:
total axial force needed to move cone w/ actuator friction = 500 lbs + 1221.7 lbs + 525.0 lbs = 2246.8 lbs
Force needed due to friction caused by mover mechanism:
The mover mechanism that moves a cone axially also applies a frictional force that resist axial movements by pushing the transmission belt against the cone. Since for the cones the angle is very shallow (20°), the force provided by the mover mechanism will always exceed the frictional resistance force due to the mover mechanism even if the coefficient of friction is 1. This might not be so for an iCVT where the angle of the variator is very steep.
Since here the force provided by the mover mechanism will always exceed the frictional resistance force due to the mover mechanism, we need to make sure that the minimum force provided by the mover mechanism will always exceed the minimum total force required. We already know that more than enough maximum force is provided by the mover mechanism to overcome the maximum “total axial force needed to move cone w/ actuator friction” from the calculations below.
The minimum total axial force needed to move cone w/ actuator friction near the neutral position = 1221.7 lbs + 525.0 lbs = 1746.8 lbs
Note: near the neutral position the axial force F for accelerating the actuator lever is not needed.
Frictional force due to 1746.8 lbs = µ * 1746.8 lbs * sin20° = 59.7 lbs
minimum total axial force needed to move cone = 1746.8 lbs + 59.7 lbs = 1806.5 lbs
The travel distance that a cone needs to be moved is:
travel distance needed = 0.33 in
Mover Mechanism for moving a cone axially
Spring to be used:
8.75” overall length, 11.54” extended length, min. force = 64.5 lbs, max. force 204 lbs
total distance spring can be travel = 11.54 in – 8.75 in = 2.79 in
Since, total distance spring can be travel / travel distance needed = 2.97 in / .33 in = 8.5,
we can use a 8 to 1 moment arm ratio between the moment arm of the spring and the moment arm of the gear of the gear rack that is used to move a cone axially. For example, if the radius of the gear is 0.5 in, then the force of the spring can be applied at 4 in long moment arm.
We can use as many springs as needed. If we use 4 springs we get:
min. force of springs = 4 * 64.5 lbs * 8 = 2064 lbs, and
max. force of spring = 4 * 204 lbs * 8 = 6528 lbs, which is more than enough.
Pneumatic actuator to be used:
We select the stroke length of the pneumatic actuator so that the moment arm of the springs and the moment arm of the pneumatic actuator are equal or almost equal, so we get:
stroke length = 3 in min.
min. force = 4 * 204 lbs = 816 lbs
Here a 31/2 in dia. cylinder will work
Compressor to be used:
cylinder speed (in/sec) = 28.8 *CFM/cylinder area
cylinder area = π * 3.5^2 = 9.62 in^2
If we use a 1.7 CFM compressor, we get:
cylinder speed (in/sec) = 28.8 *1/9.62 = 5.01 in/sec
duration to change the axial position of a cone = 3 in / 5.01 in/sec + 0.00411 sec = 0.603 sec
Power losses due to compressor:
compressor max. power consumption = V * I = 120 V * 10 A = 1200 watts = [1200 / 746] HP = 1.6 HP
When a vehicle is cruising at a steady speed, which probably about 60% of the time for average driving conditions, the compressor power consumption is 0. If a vehicle uses 50 HP on average then the CVT can be 98% efficient.
Compared to manual and automatic transmissions, a CVT 2 can provide 37% increase in fuel efficiency (link).
Regarding the losses due to the “Mover Sliding Plate Mechanism”, the “slot and pin” movement of the “Mover Sliding Plate Mechanism” is similar to a “worm gear and gear tooth movement”. Both lift an item (pin, gear tooth) using a wedge (slot, worm gear thread). This is similar to sliding/rolling an object up an incline.
Worm gear drives can be more than 90% efficient. In addition, rolling action instead of sliding action (as only available for a worm gear drive) can be used between a slot and its pin (wheel), which compresses during loading. If needed, a force twice as large as shown in the calculation above can easily provided by using a stronger actuator, stronger springs, and longer lever arms.
Also the power consumption of an iCVT should be considerably larger than that of CVT 1. An iCVT uses a variator that comprises of two pulley halves, each one identical to a steep angled cone, that each need to be moved the required axial distance. This should double the distance the variator of an iCVT has to be moved axially compared to a cone shaped like a pulley half for the same amount of transmission diameter change. And the force needed to move the variator of an iCVT should also be larger since an iCVT can have almost two half transmission belt/chain loops under tension, while a CVT 1 only has a slightly greater than half transmission belt/chain loop under tension. In addition, if iCVT uses 2 inch wide pulley halves, the chain width is 4 inches since it has to move 2 inches axially for each pulley halve, so the chain of an iCVT will be heavy. This should not be a problem at low rpm, but at high rpm the energy losses due to the momentum losses of the chain can be large.
Sample calculations 2 (preferred)
Engine Input
Maximum torque: 200 ftlbs
Maximum rpm: 6000 rpm
Use a 2:1 gearbox so that input to CVT is:
Maximum torque: 400 ftlbs
Maximum rpm: 3000 rpm
Cones
Driving Cone:
small dia. = 4.89 in (radius = 2.45 in)
large dia. = 10.15 in
angle of cone = 24 °
length of cone = [10.15  4.89 /[tan 24°]] + 1 (width of trans. belt) = [5.91 + 1] in = 6.91 in
Driven Cone:
small dia. = 4.89 in (radius = 2.45 in)
large dia. = 10.15 in
angle of cone = 24 °
length of cone = [10.15  4.89 /[tan 24°]] + 1 (width of trans. belt) = [5.91 + 1] in = 6.91 in
Transmission ratio:
Transmission ratio range available = 4.89 : 10.15 to 10.15 : 4.89 = 0.48 to 2.08 = 1 to 4.31
Teeth of cones & transmission belt
maximum torque = 400 ftlbs * [12 in/1 ft] = 4800 lbsin
maximum pulling force = 4800 lbsin / 2.45 in (small dia. of driving cone) = 1963.2 lbs
Use steel and polyamid belt for teeth, strength = 30,000 psi (lbs/in^2), therefore:
engagement area required = 1963.2 lbs / 30,000 psi = 0.0654 in^2
Use belt with cog shaped teeth with:
dia. of a tooth= .25 in
width of transmission belt = 1 in
engagement area = .25 * 1 in^2 = .25 in^2
N (factor of safety) = engagement area/ engagement area required = .25 in^2 / 0.0654 in^2 = 3.82
Regular cog belts are not as strong as shown in the calculation above. But, earlier we have shown some catalog data for a cog belt data that can transmit up to 1000 HP, which is more than the HP rating of most cars.
Axial movements required for trans. dia. change of 2 teeth
Tooth:
dia. of a tooth =.25 in
width of tooth = .375 in
Driving/Driven Cone (driven & driving cones are identical):
circum. of small dia. = π * d = 3.14 * 4.89 = 15.36 in
number of teeth at small dia. = 15.36 in / .375 in = 41 teeth
circumference of large dia. = π * d = 3.14 * 10.15 = 31.89 in
number of teeth at large dia. = 31.89 in / .375 in = 85 teeth
for a distance of 5.91 in (length – width of belt), the number of teeth increase from 41 to 85
therefore, axial distance for 1 tooth = 5.91 in / [85 teeth – 41 teeth] = 0.134 in
axial distance for 2 teeth = 0.134 in * 2 = 0.27 in
number of teeth at large dia.  number of teeth at small dia. = 85 41 = 44 teeth
Total transmission ratios available:
Total transmission ratios = [44 teeth]/”2” * “3” = 66 transmission ratios
Note: “2” since only an even number of teeth can be used for the cones, & “3” since for each axial position change 3 transmission ratios can be obtained, see “The CVT” section.
Force needed to move a cone axially due to inertia
Below is the calculation where a force F is used to move a cone of mass m:
F = m * a, hence a = F/m,
where F  axial force on cone , m  mass of cone, a  axial acceleration of cone
Lets test an axial force of 500 lbs (although this force seems large, it is manageable, please bear with us) and assume a cone and its mover mechanism weights 2 kg, then:
F = 500 lbs = 2225 N
m = 2 kg
a = F/m = 2225 N / 2 kg = 1112.5 m/s^2
In order to determine the duration that a cone can be moved axially we use:
ɵ = ω * t, from which we get t = ɵ/ω
where:
ɵ  angular rotation where the fastest cone (driven cone) can be moved
ω – maximum rpm of the fastest cone
t  duration that the fastest cone can be moved
For a 3000 rpm input, and 10.15 in dia. driving cone & a 4.89 in dia. driven cone, we get:
ω = 3000 * 10.15/4.89 rpm = 6227.0 rpm = 6395.4/[60 rot/sec] = 103.78 rot/sec = 114.9 *360°/sec = 37362.0°/sec
If we allow 170° of rotation to move the fastest cone, then we get:
t = ɵ/ω = 170° / 37362.0°/sec = 0.00455 sec
In order to determine the movement of the fastest cone due to the axial force F, we use:
s = v * t + 1/2 * a * t^2
where:
s  axial distance traveled by cone
v  initial axial velocity of cone,
a  acceleration of cone due to force F
t  duration that the axial force F can be applied
Using the values we calculated earlier we get:
s = 0 * 0.00455 sec + 1/2 * 1112.5 m/sec^2 * [0.00455 sec]^2 = 0.0115 m = (0.0094 * 39.37) in = 0.45 in
Since the amount of axial movement required is 0.27 in, and since 0.45 in > 0.27 in this will work.
Force needed to move a cone axially due to tension and friction
Force needed due to tension of transmission belt:
tension (T)= 1963.2 lbs
max. axial force needed due to axial force of T = 2 * T * tan 24° = 2 * 1963.2 * 0.45 = 1748.1 lbs
Force needed due to friction of transmission belt:
max. normal force of T = 2 * T / cos 20° = 3586.9 lbs
max. axial force needed due to friction = µ *max. normal force of T (due to belt resting on cone) + µ * T (due to tooth) = 0.1 * 3586.9 lbs + 0.1 * 1963.2lbs = 555.0 lbs
Total force needed due to transmission belt:
total axial force needed to move cone w/ actuator friction = 500 lbs + 1748.1 lbs + 555.0 lbs = 2803.1 lbs
Force needed due to friction caused by mover mechanism:
The minimum total axial force needed to move cone w/ actuator friction near the neutral position = 1748.1 lbs + 555.0 lbs = 2303.1 lbs
Note: near the neutral position the axial force F for accelerating the actuator lever is not needed.
Frictional force due to 2303.1 lbs = µ * 2303.1 lbs * sin 24° = 93.7 lbs
minimum total axial force needed to move cone = 2303.1 lbs + 93.7 lbs = 2396.8 lbs
The travel distance that a cone needs to be moved is:
travel distance needed = 0.27 in
Mover Mechanism for moving a cone axially
Spring to be used:
8.75” overall length, 11.54” extended length, min. force = 64.5 lbs, max. force 204 lbs
total distance spring can be travel = 11.54 in – 8.75 in = 2.79 in
Since, total distance spring can be travel / travel distance needed = 2.97 in / .27 in = 10,
we can use a 10 to 1 moment arm ratio between the moment arm of the spring and the moment arm of the gear of the gear rack that is used to move a cone axially. For example, if the radius of the gear is 0.5 in, then the force of the spring can be applied at 5 in long moment arm.
We can use as many springs as needed. If we use 4 springs we get:
min. force of springs = 4 * 64.5 lbs * 10 = 3870 lbs, and
max. force of spring = 4 * 204 lbs * 10 = 8160 lbs, which is more than enough.
Pneumatic actuator to be used:
We select the stroke length of the pneumatic actuator so that the moment arm of the springs and the moment arm of the pneumatic actuator are equal or almost equal, so we get:
stroke length = 3 in min.
min. force = 4 * 204 lbs = 816 lbs
Here a 31/2 in dia. cylinder will work
Compressor to be used:
cylinder speed (in/sec) = 28.8 *CFM/cylinder area
cylinder area = π * 3.5^2 = 9.62 in^2
If we use a 1.7 CFM compressor, we get:
cylinder speed (in/sec) = 28.8 *1/9.62 = 5.01 in/sec
duration to change the axial position of a cone = 3 in / 5.01 in/sec + 0.00455 sec = 0.603 sec
Power losses due to compressor:
compressor max. power consumption = V * I = 120 V * 10 A = 1200 watts = [1200 / 746] HP = 1.6 HP
If desired “Naudic iCVT Transmission Ratio Range Extender” or planetary gearbox can be used with the CVT.
If needed/desired tensioning pins can shaped/attached to the diagonally reinforced side surface of a transmission belt (see Figs. 11 & 12) directly or through the use of other means such as inserted metal pins for example. The tensioning pins shown in Figs. 11 & 12 can be used to help maintain tension (remove slack) in the reinforcement wire(s) as the transmission belt is bent downwards. Similarly tensioning pins that help maintain the tension in the reinforcement wire(s) as the transmission belt is bent upward can also be used. The proper specification for the transmission belt, the locations of the lower anchor plates and upper anchor plates, and the ideal size and positions of the tensioning pins, to satisfy the criteria of the previous paragraph and any other requirements can be obtained through experimentations (trial and error, simulations, etc) and science.
If desired, the smaller side surface or any side surface of a transmission belt, whether it is tapered or not, can also be diagonally reinforced using reinforcement wire(s) in the same manner as described in the previous paragraph. If both side surfaces of a transmission belt are diagonally reinforced, then it also needs to be ensured that the transmission belt can properly bent upwards and bent downwards as needed without twisting.
Axial movements required for trans. dia. change of 2 teeth
Tooth:
dia. of a tooth =.25 in
width of tooth = .375 in
Driving Cone:
circum. of small dia. = π * d = 3.14 * 2.86 = 8.98 in
number of teeth at small dia. = 8.98 in / .375 in = 24 teeth
circum. of large dia. = π * d = 3.14 * 7.44 = 23.25 in
number of teeth at large dia. = 23.25 in / .375 in = 62 teeth
for a distance of 6.24 in (length – width of belt), the number of teeth increase from 24 to 62
therefore, axial distance for 1 tooth = 6.24 in / [62 teeth – 24 teeth] = 0.16 in
axial distance for 2 teeth = 0.16 in * 2 = 0.33 in
number of teeth at large dia.  number of teeth at small dia. = 62 24 = 40 teeth
Driven Cone:
circum. of small dia. = π * d = 3.14 * 6.44 = 20.23 in
number of teeth at small dia. = 20.23 in / .375 in = 54 teeth
circum. of large dia. = π * d = 3.14 * 10.98 = 34.49 in
number of teeth at large dia. = 23.25 in / .375 in = 92 teeth
number of teeth at large dia.  number of teeth at small dia. = 92 54 = 40 teeth
Total transmission ratios available:
Total transmission ratios = [40 teeth (=62 – 24 or 92 54)]/”2” * “3” = 57 transmission ratios
Note: “2” since only an even number of teeth can be used for the cones, & “3” since for each axial position change 3 transmission ratios can be obtained, see “The CVT” tab.
Force needed to move a cone axially due to inertia
Below is the calculation where a force F is used to move a cone of mass m:
F = m * a, hence a = F/m,
where F  axial force on cone , m  mass of cone, a  axial acceleration of cone
Lets test an axial force of 500 lbs (although this force seems large, it is manageable, please bear with us) and assume a cone and its mover mechanism weights 2 kg, then:
F = 500 lbs = 2225 N
m = 2 kg
a = F/m = 2225 N / 2 kg = 1112.5 m/s^2
In order to determine the duration that a cone can be moved axially we use:
ɵ = ω * t, from which we get t = ɵ/ω
where:
ɵ  angular rotation where the fastest cone (driven cone) can be moved
ω – maximum rpm of the fastest cone
t  duration that the fastest cone can be moved
For a 6000 rpm input, and 7.4 in dia. driving cone & a 6.44 in dia. driven cone, we get:
ω = 6000 * 7.4/6.44 rpm = 6894.4 rpm = 6395.4/[60 rot/sec] = 114.9 rot/sec = 114.9 *360°/sec = 41366.5°/sec
If we allow 170° of rotation to move the fastest cone, then we get:
t = ɵ/ω = 170° / 41366.5°/sec = 0.00411 sec
In order to determine the movement of the fastest cone due to the axial force F, we use:
s = v * t + 1/2 * a * t^2
where:
s  axial distance traveled by cone
v  initial axial velocity of cone,
a  acceleration of cone due to force F
t  duration that the axial force F can be applied
Using the values we calculated earlier we get:
s = 0 * 0.00411 sec + 1/2 * 1112.5 m/sec^2 * [0.00411 sec]^2 = 0.0094 m = (0.0094 * 39.37) in = 0.37 in
Since the amount of axial movement required is 0.33 in, and since 0.37 in > 0.33 in this will work.
Force needed to move a cone axially due to tension and friction
Force needed due to tension of transmission belt:
tension (T)= 1678.3 lbs
max. axial force needed due to axial force of T = 2 * T * tan 20° = 2 * 1678.3 * 0.36 = 1221.7 lbs
(Note: if a Vbelt pulley, as in an iCVT, with an angle of 70° is used as then:
max. axial force needed due to axial force of T = [2 * T * tan 70°] * 1.8 = [2 * 1678.3 * 2.74] * 1.8 = 16600.1 lbs.
The multiplication of 1.8 is used because an iCVT can have almost two half transmission belt/chain loops under tension, while a CVT 1 only has a slightly greater than half transmission belt/chain loop under tension.)
Force needed due to friction of transmission belt:
max. normal force of T = 2 * T / cos 20° = 3572.1 lbs
(Note: if a Vbelt pulley, as in an iCVT, with an angle of 70° is used as then:
max. normal force of T = [2 * T / cos 70°] * 1.8 = 9814.2 lbs * 1.8 = 17665.6 lbs
The multiplication of 1.8 is used because an iCVT can have almost two half transmission belt/chain loops under tension, while a CVT 1 only has a slightly greater than half transmission belt/chain loop under tension.)
max. axial force needed due to friction = µ *max. normal force of T (due to belt resting on cone) + µ * T (due to tooth) = 0.1 * 3572.1 lbs + 0.1 * 1678.3 lbs = 525.0 lbs
Total force needed due to transmission belt:
total axial force needed to move cone w/ actuator friction = 500 lbs + 1221.7 lbs + 525.0 lbs = 2246.8 lbs
Force needed due to friction caused by mover mechanism:
The mover mechanism that moves a cone axially also applies a frictional force that resist axial movements by pushing the transmission belt against the cone. Since for the cones the angle is very shallow (20°), the force provided by the mover mechanism will always exceed the frictional resistance force due to the mover mechanism even if the coefficient of friction is 1. This might not be so for an iCVT where the angle of the variator is very steep.
Since here the force provided by the mover mechanism will always exceed the frictional resistance force due to the mover mechanism, we need to make sure that the minimum force provided by the mover mechanism will always exceed the minimum total force required. We already know that more than enough maximum force is provided by the mover mechanism to overcome the maximum “total axial force needed to move cone w/ actuator friction” from the calculations below.
The minimum total axial force needed to move cone w/ actuator friction near the neutral position = 1221.7 lbs + 525.0 lbs = 1746.8 lbs
Note: near the neutral position the axial force F for accelerating the actuator lever is not needed.
Frictional force due to 1746.8 lbs = µ * 1746.8 lbs * sin20° = 59.7 lbs
minimum total axial force needed to move cone = 1746.8 lbs + 59.7 lbs = 1806.5 lbs
The travel distance that a cone needs to be moved is:
travel distance needed = 0.33 in
Mover Mechanism for moving a cone axially
Spring to be used:
8.75” overall length, 11.54” extended length, min. force = 64.5 lbs, max. force 204 lbs
total distance spring can be travel = 11.54 in – 8.75 in = 2.79 in
Since, total distance spring can be travel / travel distance needed = 2.97 in / .33 in = 8.5,
we can use a 8 to 1 moment arm ratio between the moment arm of the spring and the moment arm of the gear of the gear rack that is used to move a cone axially. For example, if the radius of the gear is 0.5 in, then the force of the spring can be applied at 4 in long moment arm.
We can use as many springs as needed. If we use 4 springs we get:
min. force of springs = 4 * 64.5 lbs * 8 = 2064 lbs, and
max. force of spring = 4 * 204 lbs * 8 = 6528 lbs, which is more than enough.
Pneumatic actuator to be used:
We select the stroke length of the pneumatic actuator so that the moment arm of the springs and the moment arm of the pneumatic actuator are equal or almost equal, so we get:
stroke length = 3 in min.
min. force = 4 * 204 lbs = 816 lbs
Here a 31/2 in dia. cylinder will work
Compressor to be used:
cylinder speed (in/sec) = 28.8 *CFM/cylinder area
cylinder area = π * 3.5^2 = 9.62 in^2
If we use a 1.7 CFM compressor, we get:
cylinder speed (in/sec) = 28.8 *1/9.62 = 5.01 in/sec
duration to change the axial position of a cone = 3 in / 5.01 in/sec + 0.00411 sec = 0.603 sec
Power losses due to compressor:
compressor max. power consumption = V * I = 120 V * 10 A = 1200 watts = [1200 / 746] HP = 1.6 HP
When a vehicle is cruising at a steady speed, which probably about 60% of the time for average driving conditions, the compressor power consumption is 0. If a vehicle uses 50 HP on average then the CVT can be 98% efficient.
Compared to manual and automatic transmissions, a CVT 2 can provide 37% increase in fuel efficiency (link).
Regarding the losses due to the “Mover Sliding Plate Mechanism”, the “slot and pin” movement of the “Mover Sliding Plate Mechanism” is similar to a “worm gear and gear tooth movement”. Both lift an item (pin, gear tooth) using a wedge (slot, worm gear thread). This is similar to sliding/rolling an object up an incline.
Worm gear drives can be more than 90% efficient. In addition, rolling action instead of sliding action (as only available for a worm gear drive) can be used between a slot and its pin (wheel), which compresses during loading. If needed, a force twice as large as shown in the calculation above can easily provided by using a stronger actuator, stronger springs, and longer lever arms.
Also the power consumption of an iCVT should be considerably larger than that of CVT 1. An iCVT uses a variator that comprises of two pulley halves, each one identical to a steep angled cone, that each need to be moved the required axial distance. This should double the distance the variator of an iCVT has to be moved axially compared to a cone shaped like a pulley half for the same amount of transmission diameter change. And the force needed to move the variator of an iCVT should also be larger since an iCVT can have almost two half transmission belt/chain loops under tension, while a CVT 1 only has a slightly greater than half transmission belt/chain loop under tension. In addition, if iCVT uses 2 inch wide pulley halves, the chain width is 4 inches since it has to move 2 inches axially for each pulley halve, so the chain of an iCVT will be heavy. This should not be a problem at low rpm, but at high rpm the energy losses due to the momentum losses of the chain can be large.
Sample calculations 2 (preferred)
Engine Input
Maximum torque: 200 ftlbs
Maximum rpm: 6000 rpm
Use a 2:1 gearbox so that input to CVT is:
Maximum torque: 400 ftlbs
Maximum rpm: 3000 rpm
Cones
Driving Cone:
small dia. = 4.89 in (radius = 2.45 in)
large dia. = 10.15 in
angle of cone = 24 °
length of cone = [10.15  4.89 /[tan 24°]] + 1 (width of trans. belt) = [5.91 + 1] in = 6.91 in
Driven Cone:
small dia. = 4.89 in (radius = 2.45 in)
large dia. = 10.15 in
angle of cone = 24 °
length of cone = [10.15  4.89 /[tan 24°]] + 1 (width of trans. belt) = [5.91 + 1] in = 6.91 in
Transmission ratio:
Transmission ratio range available = 4.89 : 10.15 to 10.15 : 4.89 = 0.48 to 2.08 = 1 to 4.31
Teeth of cones & transmission belt
maximum torque = 400 ftlbs * [12 in/1 ft] = 4800 lbsin
maximum pulling force = 4800 lbsin / 2.45 in (small dia. of driving cone) = 1963.2 lbs
Use steel and polyamid belt for teeth, strength = 30,000 psi (lbs/in^2), therefore:
engagement area required = 1963.2 lbs / 30,000 psi = 0.0654 in^2
Use belt with cog shaped teeth with:
dia. of a tooth= .25 in
width of transmission belt = 1 in
engagement area = .25 * 1 in^2 = .25 in^2
N (factor of safety) = engagement area/ engagement area required = .25 in^2 / 0.0654 in^2 = 3.82
Regular cog belts are not as strong as shown in the calculation above. But, earlier we have shown some catalog data for a cog belt data that can transmit up to 1000 HP, which is more than the HP rating of most cars.
Axial movements required for trans. dia. change of 2 teeth
Tooth:
dia. of a tooth =.25 in
width of tooth = .375 in
Driving/Driven Cone (driven & driving cones are identical):
circum. of small dia. = π * d = 3.14 * 4.89 = 15.36 in
number of teeth at small dia. = 15.36 in / .375 in = 41 teeth
circumference of large dia. = π * d = 3.14 * 10.15 = 31.89 in
number of teeth at large dia. = 31.89 in / .375 in = 85 teeth
for a distance of 5.91 in (length – width of belt), the number of teeth increase from 41 to 85
therefore, axial distance for 1 tooth = 5.91 in / [85 teeth – 41 teeth] = 0.134 in
axial distance for 2 teeth = 0.134 in * 2 = 0.27 in
number of teeth at large dia.  number of teeth at small dia. = 85 41 = 44 teeth
Total transmission ratios available:
Total transmission ratios = [44 teeth]/”2” * “3” = 66 transmission ratios
Note: “2” since only an even number of teeth can be used for the cones, & “3” since for each axial position change 3 transmission ratios can be obtained, see “The CVT” section.
Force needed to move a cone axially due to inertia
Below is the calculation where a force F is used to move a cone of mass m:
F = m * a, hence a = F/m,
where F  axial force on cone , m  mass of cone, a  axial acceleration of cone
Lets test an axial force of 500 lbs (although this force seems large, it is manageable, please bear with us) and assume a cone and its mover mechanism weights 2 kg, then:
F = 500 lbs = 2225 N
m = 2 kg
a = F/m = 2225 N / 2 kg = 1112.5 m/s^2
In order to determine the duration that a cone can be moved axially we use:
ɵ = ω * t, from which we get t = ɵ/ω
where:
ɵ  angular rotation where the fastest cone (driven cone) can be moved
ω – maximum rpm of the fastest cone
t  duration that the fastest cone can be moved
For a 3000 rpm input, and 10.15 in dia. driving cone & a 4.89 in dia. driven cone, we get:
ω = 3000 * 10.15/4.89 rpm = 6227.0 rpm = 6395.4/[60 rot/sec] = 103.78 rot/sec = 114.9 *360°/sec = 37362.0°/sec
If we allow 170° of rotation to move the fastest cone, then we get:
t = ɵ/ω = 170° / 37362.0°/sec = 0.00455 sec
In order to determine the movement of the fastest cone due to the axial force F, we use:
s = v * t + 1/2 * a * t^2
where:
s  axial distance traveled by cone
v  initial axial velocity of cone,
a  acceleration of cone due to force F
t  duration that the axial force F can be applied
Using the values we calculated earlier we get:
s = 0 * 0.00455 sec + 1/2 * 1112.5 m/sec^2 * [0.00455 sec]^2 = 0.0115 m = (0.0094 * 39.37) in = 0.45 in
Since the amount of axial movement required is 0.27 in, and since 0.45 in > 0.27 in this will work.
Force needed to move a cone axially due to tension and friction
Force needed due to tension of transmission belt:
tension (T)= 1963.2 lbs
max. axial force needed due to axial force of T = 2 * T * tan 24° = 2 * 1963.2 * 0.45 = 1748.1 lbs
Force needed due to friction of transmission belt:
max. normal force of T = 2 * T / cos 20° = 3586.9 lbs
max. axial force needed due to friction = µ *max. normal force of T (due to belt resting on cone) + µ * T (due to tooth) = 0.1 * 3586.9 lbs + 0.1 * 1963.2lbs = 555.0 lbs
Total force needed due to transmission belt:
total axial force needed to move cone w/ actuator friction = 500 lbs + 1748.1 lbs + 555.0 lbs = 2803.1 lbs
Force needed due to friction caused by mover mechanism:
The minimum total axial force needed to move cone w/ actuator friction near the neutral position = 1748.1 lbs + 555.0 lbs = 2303.1 lbs
Note: near the neutral position the axial force F for accelerating the actuator lever is not needed.
Frictional force due to 2303.1 lbs = µ * 2303.1 lbs * sin 24° = 93.7 lbs
minimum total axial force needed to move cone = 2303.1 lbs + 93.7 lbs = 2396.8 lbs
The travel distance that a cone needs to be moved is:
travel distance needed = 0.27 in
Mover Mechanism for moving a cone axially
Spring to be used:
8.75” overall length, 11.54” extended length, min. force = 64.5 lbs, max. force 204 lbs
total distance spring can be travel = 11.54 in – 8.75 in = 2.79 in
Since, total distance spring can be travel / travel distance needed = 2.97 in / .27 in = 10,
we can use a 10 to 1 moment arm ratio between the moment arm of the spring and the moment arm of the gear of the gear rack that is used to move a cone axially. For example, if the radius of the gear is 0.5 in, then the force of the spring can be applied at 5 in long moment arm.
We can use as many springs as needed. If we use 4 springs we get:
min. force of springs = 4 * 64.5 lbs * 10 = 3870 lbs, and
max. force of spring = 4 * 204 lbs * 10 = 8160 lbs, which is more than enough.
Pneumatic actuator to be used:
We select the stroke length of the pneumatic actuator so that the moment arm of the springs and the moment arm of the pneumatic actuator are equal or almost equal, so we get:
stroke length = 3 in min.
min. force = 4 * 204 lbs = 816 lbs
Here a 31/2 in dia. cylinder will work
Compressor to be used:
cylinder speed (in/sec) = 28.8 *CFM/cylinder area
cylinder area = π * 3.5^2 = 9.62 in^2
If we use a 1.7 CFM compressor, we get:
cylinder speed (in/sec) = 28.8 *1/9.62 = 5.01 in/sec
duration to change the axial position of a cone = 3 in / 5.01 in/sec + 0.00455 sec = 0.603 sec
Power losses due to compressor:
compressor max. power consumption = V * I = 120 V * 10 A = 1200 watts = [1200 / 746] HP = 1.6 HP
If desired “Naudic iCVT Transmission Ratio Range Extender” or planetary gearbox can be used with the CVT.