**Notes:**
1.
2. |
The calculations below are for a CVT 1, and not our preferred CVT 4. But the calculations are also relevant for a CVT 4, since a CVT 4 is similar to a CVT 1.
All catalog parts are shown in the "Cost & Partslist" tab |

**Sample Calculations 1**

**Engine Input**

Maximum torque: 200 ft-lbs

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 ft-lbs * [12 in/1 ft] = 2400 lbs-in

maximum pulling force = 2400 lbs-in / 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

**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 V-belt 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 V-belt 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 3-1/2 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 ft-lbs

Maximum rpm: 6000 rpm

Use a 2:1 gearbox so that input to CVT is:

Maximum torque: 400 ft-lbs

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 ft-lbs * [12 in/1 ft] = 4800 lbs-in

maximum pulling force = 4800 lbs-in / 2.45 in (small dia. of driving cone) = 1963.2 lbs

Please see transmission belt for Sample Calculation 1

**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 3-1/2 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