Detailed Calculation for a CVT 2 (Work in Progress)
Note:
Calculation is for a CVT 2 described under the "Economical CVT 2" tab.
After working on the calculations for the rotational position adjustment system for a transmission pulley we found it impractical, despite having made additional improvements. We found it more demanding to make it work than the transmission ratio changing actuator, since it is also spinning at 6000 rpm where we have to take into consideration large centrifugal forces and vibrations and we also have to take into consideration space limitations.
The distance that needs to be traveled by the transmission ratio changing actuator is also less than that of the transmission belts.
Instead of having to deal with two fast distance changing mechanisms, which can generate shock loads, we decided to use only one. This will simplify the design as to reduce cost and increase reliability. Based on our calculations (item 3 of this tab), we don't see a problem.
Calculation is for a CVT 2 described under the "Economical CVT 2" tab.
After working on the calculations for the rotational position adjustment system for a transmission pulley we found it impractical, despite having made additional improvements. We found it more demanding to make it work than the transmission ratio changing actuator, since it is also spinning at 6000 rpm where we have to take into consideration large centrifugal forces and vibrations and we also have to take into consideration space limitations.
The distance that needs to be traveled by the transmission ratio changing actuator is also less than that of the transmission belts.
Instead of having to deal with two fast distance changing mechanisms, which can generate shock loads, we decided to use only one. This will simplify the design as to reduce cost and increase reliability. Based on our calculations (item 3 of this tab), we don't see a problem.
1.

Engine Input
Maximum torque: 200 ftlbs Maximum rpm: 6000 rpm 1st ratio (torque input : torque output) of car: 1 : 2.5 Note: Specs. for engines can be found here, link. 
2. 
CVT 2 Note: For cars space is limited, for ships, generators, trucks, customized cars, tanks, drilling rigs, conveyer belts, windmills, tractors, etc., space is less of a concern. In order to be able to attach a CVT 2 under the floor of a car so that it can be easily retrofitted to existing cars, we use a "transmission ratio range extender" of an iCVT (link). Here for the first range the input shaft drives the output shaft, and for the second range the output shaft drives the input shaft. A "transmission ratio range extender" allows us to use cones that have a larger dia. of only 5 in. Even with a "transmission ratio range extender", our CVT should not exceed the target range of $2000 to $5000. Two single tooth cones and two transmission pulleys cost probably less than $200. This leaves plenty of room as far as cost is concerned in order not to exceed the target range of $2000 to $5000. A hybrid cost about $8,000 more than a regular car, and it does not offer an increase in acceleration that a CVT 2 can provide. A regular 7speed transmission can cost more then $3000, which means that the improvement in performance provide by a CVT 2 cost only $2,000 if it was priced at $5,000. If no "transmission ratio range extender" is used, then the following is relevant. The approximate dimension of an engine compartment of a Honda Civic is shown below: height: 26 in (there is enough room for 10 in dia. cones) length: 30 in (if 10 in dia. cones & 2 in dia. transmission pulleys are used, lenght of CVT 2 is about 20 in) width: 67 in (if 10 in dia. cones are used, width of CVT 2 is about 26 in) Room can be made in an engine compartment to accommodate a CVT 2. In real wheel drive cars, a CVT 2 can be easily placed in the trunk. Here a drive shaft can be used to couple the engine up front with the CVT 2 in the trunk, and then the CVT 2 can be easily couple to the rear wheels using gears. If a CVT 2 can provide 30% increase in fuel efficiency (37% according to this link), and allows a $30,000 car to outperform a $90,000 car due to the better acceleration than having to accommodate a CVT 2 should not be an issue. 
a. 
Cones & Pulleys Dimensions: Single Tooth Cones (cones): small dia. = 2 in (radius = 1 in) large dia. = 2 in * 5 (for 5:1 (large dia. : small dia.) ratio)/2 (since use a 2 range extender) = 5 in Transmission Pulleys (pulleys): dia. = 2 in. 
b.

Teeth of Single Tooth Cones & Transmission Belt
Maximum pulling load in transmission belts (belts) maximum torque = 200 ftlbs = 200 ftlbs * (12 in /1 ft) = 2400 inlbs smallest radius where torque is applied = 1 in (= radius of trans. pulleys & radius of smallest end of cones) maximum pulling force = maximum torque / smallest radius = 2400 inlbs / 1 in = 2400 lbs Material & size for transmission belt use steel and polyamid belt (Polyamid strength, Polyamid Belt), strength = 30,000 psi (lbs/in^2) engagement area required = 2400 lbs / 30,000 psi = 0.08 in^2 use half circular shaped cog shaped teeth with radius of: .25 in use width of transmission belt of: .625 in engagement area = .25 in * .625 in = .156 in^2 N (factor of safety) = engagement area / engagement area required = .156 in^2 / 0.08 in^2 = 1.95 Notes:  Transmission belts can be reinforced with kevlar or steel reinforcement to reduce stretching.  A chain or block transmission belt can be used to eliminate the need for pretension in the belts.  Support pulleys that also help maintain the axial alignment of the transmission belts can be positioned around the circumference of the cones to eliminate the need for large pretension in the belts since they can prevent slack from forming on the portion of the transmission belts that are in contact with their cone. The support pulleys can be positioned using slider and slides (preferable), or springs. Pulley tooth width, looking at Fig. 4 at the "Business Plan,..." tab, the width of a pulley tooth is about 7/5 times the depth of a belt tooth (it is easier to measure on a straight portion of the belt), so its safe to use: width of a pulley & single tooth cone tooth = 7/5 * dept of belt tooth = 7/5 * .25 in = .35 in 
c.

Transmission ratios:
First transmission ratio range: Setup: input shaft  pulleys I output shaft  cones Ratios (input : output) = 2 in (dia. of pulleys) : 5 in (large dia. of cones) = 1 : 2.5 2 in (dia. of pulleys) : 2 in (small dia. of cones) = 1 : 1 Second transmission ratio range: Setup: input shaft  cones I output shaft  pulleys Ratios (input : output) = 2 in (small dia. of cones): 2 in (dia. of pulleys) = 1 : 1 5 in (large dia. of cones): 2 in (dia. of pulleys) = 2.5 : 1 
d.


3.

Transmission Ratio Changing Actuator
The transmission ratio is only changed when no torque is transmit. First we ignore the resistance to transmission ratio change due to the pretension in the transmission belt. This additional force can be added later. 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 100 lbs and assume a cone weights 0.5 kg, then: F = 100 lbs = 445 N m = 0.5 kg a = F/m = 445 N / 0.5 kg = 890 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 adjustment can be provided, ω  rpm of single tooth cones, t  duration that the braking force can be applied. ω = 6000 rpm = 100 rot/sec = 36000°/sec; assuming a 6000 rpm engine, and if we allow 170° of rotation to provide adjustment, then from ɵ = ω * t, we get t = ɵ/ω = 170° / 36000°/sec = 0.00472 sec In order to determine the movement of the transmission belt due to the braking torque, 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 can be applied using the values we calculated earlier we get: s = 0 * 0.0047 sec + 1/2 * 890 m/sec^2 * (0.0047 sec)^2 = 0.00992 m = (0.00992 * 39.37) in = 0.391 in since the amount of axial movement required is 0.096 in (see item 2d of this tab), and since 0.391 in > 0.096 in this will work. Shock loads due to axial movements of the cones From iterations we found the minimum angle of rotation to provide adjustment for an axial force on a cone of 100 lbs is 85°. From iterations we found that the minimum axial force on a cone, F, for 170° of rotation to provide adjustment is only 25 lbs. And here the minimum acceleration is 222 m/s^2. For the minimum acceleration of 222 m/s^2, the maximum velocity of the cone is: v = a * t = 222 m/s^2 * 0.00472 s = 1.05 m/s If this seems in error, keep in mind that 0.096 in is only 0.096 * 2.54 / 100 m = 0.00243 m. And 1.05 m/s * t = 1.05 m/s * 0.00472 s = 0.005 m. In order to get an idea of the stopping shock loads felt during the axial position change of a cone we dropped a metal cup from a height of 1 meter. Here it took less 1 second for the metal cup to hit the ground, the final velocity of the metal cup is probably larger than 1.05 m/s, since it's initial velocity is 0 (probably about twice as large). Unless the cones are very heavy we don't see a problem. If desired a larger initial force and some final damping can be used to reduce the stopping shock loads. If the tension in the transmission belts are very large and the angles of the cones are very steep, than loading shock loads, which correspond to the axial force needed to move the transmission belts, exist during the initial application of an axial force, but this can be avoided here. Applying a 25 lbs force to cover a distance of 0.096 in instead of a 13 lbs force to cover a distance of 0.096 in / 2 = 0.048 in, is not worth having to complicated the system using rotational position adjusters, which can generate shock loads themselves that can be larger and more difficult to damp. The stopping shock loads due to this force can be damped, and the loading shock loads mainly depend on the axial force needed to move the transmission belts and not this force. Avoiding large pretension in the transmission belts Large pretension in the transmission belts can be avoided by using support pulleys that push/hold the transmission belts toward the surface of their cone around the section of the transmission belts that is covering the cones. Here the support pulleys will remove any slack at the section of the belts that is covering the cones, and prevent an engaged tooth of a transmission belt to loose engagement with the tooth of its cone. Large pretension in a transmission belt can also be avoided by using a support pulleys (see Fig. 1A of "Introduction" tab) that is driven to rotate with its transmission belt and that engages its transmission belt for torque transmission. This setup allows minimum slip between the transmission belts and its support pulley and this will prevent any slack at the section of the transmission belt that is covering its cone. In case large pretension is required, a "cone with 1 sliding tooth", which is an alternate embodiment of the invention, can be used. A "cone with 1 sliding tooth" can be used with a chain so that tensioners are no needed. A "cone with 1 sliding tooth" is shown in Figs. 2, 3, & 4 below. A "cone with 1 sliding tooth" is heavier, more complicated, and more expensive then a single tooth cone. But, if large pretension using single tooth cones is unavoidable, a "cone with 1 sliding tooth" will provide an alternate option. Fig. 2: Sectional views of a "Cone with 1 sliding tooth"
Fig. 3: Spline of a CVT 2 where two "Cone with 1 sliding tooth" are mounted
Fig. 4: Sideview of a CVT 2 using two "Cone with 1 sliding tooth"
