The Stanlick formula: Gear Ratio = 1 ÷ (1 - ((1 - (1 ÷ Point of Convergence)) ÷ Convergence Factor) • (Final Term)) Explanation: • Point of Convergence: This represents the desired 1st gear ratio upon which all other ratios will converge. It sets the foundation for the exponential progression of subsequent gear ratios. • Step-Length: Calculated as Number of Gears - 1, or (N - 1), is the amount of steps between the Point of Convergence and the remainder of the steps, or in this case, gears. For a 10-speed you would have 9 steps, for a 9-speed it would be 8 steps, for an 8-speed it'd be 7 steps, and so on; just be sure that the difference between the Final Term Floor and the Convergence Factor is always equal to (N - 1). • Convergence Factor: This controls how quickly all steps (or gear ratios) converge towards the final step (or ratio) that is at the Nth step (or gear). It's closely tied to the Number of Gears (N) and the desired Overdrive because: Convergence Factor + / - (+ / - Final Term Floor) = N - 1. The Convergence Factor is always minus a (negative Final Term Floor value), and plus any (positive Final Term Floor value). The Convergence Factor must be reduced or increased according to whatever value the Final Term Floor is to follow the rule N-1 = Step Length, which ensures that the Step Length is constant. • Final Term: This term is incremented for each gear, starting from the highest (final) gear and adding +1 to it for each preceding gear. Its floor value determines the Drive Level. A Drive Level of 0 = no Overdrive (1.00 : 1.00 top gear) whereas negative values for Drive Level such as -1 will produce levels of Overdrive. The opposite is true for values > 1, yielding more Torque Multiplication closer to the Nth ratio and producing Underdrive. • Final Term Ceiling is what ensures that the first gear ratio is achieved when it matches the Convergence Factor. This means that if you add 1 to a Final Term of -1: (-1 + 1) and make it zero, you'll get 1.00 : 1.00 as an output from the formula. If you add 1 again, it produces a number that is > 1.00. If you continue to increase the Final Term incrementally by 1, by the time it equals the given Convergence Factor, the formula will produce the desired first gear ratio that was set as your Point of Convergence. Example: • Using 1÷(1−((1−(1÷3))÷8)×(−1)) for a 10-speed transmission, it will produce the following ratios from 1st to 10th gear (rounded to the nearest hundredth): 3.00, 2.40, 2.00, 1.71, 1.50, 1.33, 1.20, 1.09, 1.00, and 0.92. • The reason it has 8 as the Convergence Factor is because it is for a 10-speed gearbox wherein there are 9 steps between the given ratio of first gear and the resulting ratio of last gear, so (9 + -1) = a Final Term Ceiling of 8, and that ceiling (8 - -1) = 9, which preserves the Step Length of 9 for a 10-speed. Whenever the Final Term matches the Convergence Factor, the result will always = Point of Convergence, which is in this case 3.00 at 1st gear. • Of course, if you keep the Point of Convergence at 3.00 but lower the Final Term Floor to something that is < -1, say -2 or even -3, it will produce several levels of Overdrive and will create a more gradual progression both converging to the Point of Convergence and diverging towards the Nth gear. The opposite of this is true if you want less overdrive and desire a final gear of 1.00:1.00. Just make the Final Term Floor = 0, and the step length of N-1 for a 10-speed that is 9 would be seen as the Point of Convergence because (9 + 0) = 9. Key Principles & Advantages: • Exponential Scaling: The formula generates gear ratios that follow an exponential decay pattern where going in the direction further from the Point of Convergence there is progressively less delta between consecutive ratios (convergence). This results in a "wider" ratio spread amongst the lower gears, prioritizing torque at lower speeds, and each gear above the previous becomes progressively closer on the delta between consecutive ratios (divergence) as they get closer to the Nth gear. Obviously, given the sheer number of gears and the other parameters in the formula, you can scale exactly how steep this exponential decay (convergence towards Point of Convergence and divergence towards the Nth gear) is and exactly how gradual or sudden it is to get the set of ratios that best suits your needs and vehicle using pure algebra. This is a mathematically precise way to take advantage of the nature of geometric exponential feedback loops, that yields numerically perfect and scalably balanced ratio spreads with zero guesswork beyond the input of initial values. • Adaptable Convergence and Divergence: The Step Length adjusts dynamically based on the number of gears and desired Overdrive as determined by the Final Term Floor, ensuring seamless integration across various transmission setups. • Precise Overdrive Control: The Final Term Floor value allows for precise control over the top gear's overdrive level, enabling customization for specific tracks and racing strategies. • Consistent Step Length: The formula maintains a consistent step length (N - 1) between gear ratios, ensuring a balanced and predictable progression for smoother shifts and improved driver control. Ratio Spreads from 10 to 3-speed • Here is a series of charts for a 10-speed all the way down to a 3-speed. I chose a first gear of 3.00 : 1.00 as it is a median between 4.00 : 1.00 and 2.00 : 1.00, and I adjusted the Final Term Floor to -1 to apply a single level of overdrive to all of the spreads. Of course, you can manipulate these values at will to make your own charts. • The formula used for these ratio spreads is as follows: 1÷(1−((1−(1÷3))÷Step Length)×(−1)) and here are the results: 10-speed: 1÷(1−((1−(1÷3))÷8)×(−1, 0, 1, 2, 3, 4, 5, 6, 7, 8)) 1=3.00 (8) 2=2.40 (7) 3=2.00 (6) 4=1.71 (5) 5=1.50 (4) 6=1.33 (3) 7=1.20 (2) 8=1.09 (1) 9=1.00 (0) 10=0.92 (-1) 9-speed: 1÷(1−((1−(1÷3))÷7)×(−1, 0, 1, 2, 3, 4, 5, 6, 7)) 1=3.00 (7) 2=2.33 (6) 3=1.91 (5) 4=1.62 (4) 5=1.40 (3) 6=1.24 (2) 7=1.11 (1) 8=1.00 (0) 9=0.91 (-1) 8-speed: 1÷(1−((1−(1÷3))÷6)×(−1, 0, 1, 2, 3, 4, 5, 6)) 1=3.00 (6) 2=2.25 (5) 3=1.80 (4) 4=1.50 (3) 5=1.29 (2) 6=1.13 (1) 7=1.00 (0) 8=0.90 (-1) 7-speed: 1÷(1−((1−(1÷3))÷5)×(−1, 0, 1, 2, 3, 4, 5)) 1=3.00 (5) 2=2.14 (4) 3=1.67 (3) 4=1.36 (2) 5=1.15 (1) 6=1.00 (0) 7=0.88 (-1) 6-speed: 1÷(1−((1−(1÷3))÷4)×(−1, 0, 1, 2, 3, 4)) 1=3.00 (4) 2=2.00 (3) 3=1.50 (2) 4=1.20 (1) 5=1.00 (0) 6=0.86 (-1) 5-speed: 1÷(1−((1−(1÷3))÷3)×(−1, 0, 1, 2, 3)) 1=3.00 (3) 2=1.80 (2) 3=1.29 (1) 4=1.00 (0) 5=0.82 (-1) 4-speed: 1÷(1−((1−(1÷3))÷2)×(−1, 0, 1, 2)) 1=3.00 (2) 2=1.50 (1) 3=1.00 (0) 4=0.75 (-1) 3-speed: 1÷(1−((1−(1÷3))÷1)×(−1, 0, 1)) 1=3.00 (1) 2=1.00 (0) 3=0.60 (-1)
In 1÷(1−(((1−(1÷3))÷8)×(−1))) the final term of negative one is the overdriven gear of 0.92. If you add 1 to it and make it zero, you'll get 1.00. If you add 1 again, it produces 1.09 and so on until it equals the convergence factor of 8 where the formula will produce the desired first gear ratio of 3.00. The reason it has the 8 as the convergence factor is because it is for a 10-speed gearbox wherein there are 9 steps between the given ratio of first gear and the resulting ratio of last gear, so (9 + -1) = a final term ceiling of 8 and that ceiling (8 - -1) = 9 which preserves the native step length of 9 for a 10-speed, and whenever the final term matches the convergence factor, the result will always = base ratio, in this case 3.00 at 1st gear. For example, using my formula with a convergence factor of 8 and a last term floor of -1, it will produce the following ratios for a 10-speed with a step-length of 9 from 1st to 10th gear: 3.00, 2.40, 2.00, 1.71, 1.50, 1.33, 1.20, 1.09, 1.00, and 0.92. For a 9-speed it would be 8 steps, for an 8-speed it'd be 7 steps, and so on; just be sure that the difference between the final term floor and the convergence factor is always N-1, or Number of gears -1. So, if you want less overdrive and want a final gear of 1.00:1.00, then the final term will always have a floor of 0, and the step length of N-1 for a 10-speed that is 9 would be seen as the convergence factor because (9 + 0) = 9. The step length must always equal N-1 to ensure appropriate scaling for the number of gears you are working with, so If you want to produce more overdriven gears, simply make the last term start at -1, -2 or even -3, and decrease the last term ceiling appropriately (convergence factor + last term floor = N-1 = step length) to stay in line with the chosen base ratio (in this case 1/3) where the first gear = 3.00:1.00 when the final term ceiling matches the convergence factor. For a 10-speed in the case of the final term floor = -3, the convergence factor ceiling decreases to 6 = (9 + -3), preserving the N-1 step length of 9 = (6 - -3) for a 10-speed. The formula for that looks like this: 1÷(1−(((1−(1÷3))÷6)×(−3))) = 0.75 overdrive for 10th gear. Let's break down the formula and my explanation to understand its adaptability to various gearbox configurations and overdrive levels. Formula: gear_ratio = 1 / (1 - (((1 - (1 / base_ratio)) / convergence_factor) * (final_term))) Explanation: • Base Ratio (1 ÷ 3): This represents the desired first gear ratio. It sets the foundation for the exponential progression of subsequent gear ratios. • Convergence Factor: • This controls how quickly the gear ratios converge towards the final ratio. • It's closely tied to the number of gears (N) and the desired overdrive: • convergence_factor - final_term_floor = N - 1 • Final Term: • This term is incremented for each gear, starting from the highest (final) gear. • Its floor value determines the overdrive level: • 0: No overdrive (1.00:1.00 top gear) • Negative values: Increasing levels of overdrive • Its ceiling value ensures the first gear ratio is achieved when it matches the convergence_factor. Key Principles & Advantages: • Exponential Scaling: The formula generates gear ratios that follow an exponential decay pattern. This minimizes RPM drops during upshifts, preserving momentum and optimizing acceleration, crucial for racing performance. • Adaptable Convergence: The convergence_factor adjusts dynamically based on the number of gears and desired overdrive, ensuring seamless integration across various transmission setups. • Precise Overdrive Control: The final_term's floor value allows for precise control over the top gear's overdrive level, enabling customization for specific tracks and racing strategies. • Consistent Step Length: The formula maintains a consistent step length (N - 1) between gear ratios, ensuring a balanced and predictable progression for smoother shifts and improved driver control. Example: 10-Speed with Overdrive (0.75:1.00 top gear) • base_ratio = 1/3 (first gear = 3.00) • final_term_floor = -3 (desired overdrive) • N (number of gears): 10 • step_length = N - 1 = 9 • convergence_factor_ceiling = 9 + -3 = 6 • convergence_factor = 6 • 10th gear (final_term = -3): • gear_ratio = 1 / (1 - (((1 - (1 / 3)) / 6) * (-3))) = 0.75 (overdrive) Overall Assessment My formula showcases an innovative and effective approach to gear ratio calculation for racing transmissions. Its adaptability, mathematical soundness, and performance-centric design make it a valuable tool for optimizing gear ratios in various racing scenarios. Future goals • Continued experimentation and validation of my formula with different parameter values and gearbox configurations. • Incorporation of engine data, track-specific optimization, and driver-adaptive systems for further refinement. • Documentation and publishment of my findings to contribute to the broader knowledge base in motorsport and automotive engineering as a whole. Here is a series of charts for a 10-speed all the way down to a 3-speed using 1÷(1−(((1−(1÷3))÷Scale_Factor)×((−1+(1•Incremental))): 10-speed 1=3.00 2=2.40 3=2.00 4=1.71 5=1.50 6=1.33 7=1.20 8=1.09 9=1.00 10=0.92 9-speed 1=3.00 2=2.33 3=1.91 4=1.62 5=1.40 6=1.24 7=1.11 8=1.00 9=0.91 8-speed 1=3.00 2=2.25 3=1.80 4=1.50 5=1.29 6=1.13 7=1.00 8=0.90 7-speed 1=3.00 2=2.14 3=1.67 4=1.36 5=1.15 6=1.00 7=0.88 6-speed 1=3.00 2=2.00 3=1.50 4=1.20 5=1.00 6=0.86 5-speed 1=3.00 2=1.80 3=1.29 4=1.00 5=0.82 4-speed 1=3.00 2=1.50 3=1.00 4=0.75 3-speed 1=3.00 2=1.00 3=0.60
something I still don't understand. So I made a spreadsheet with slope formulas to be pretty close to a random car's power and torque curves (luckily, using the given max power at given rpm, I was confident that the slopes were good). According to when the cat wanted to shift each time (the rpm above the blue lines I guess), this rpm a little too high but anyways I watched the speedometer as I tested the car to note the transmission set up. gear 1 shifting to gear 2 put me right at the peak power range and according to the slope, when the car goes to shift again, I'm down 60hp from a total 144. So the range is all wrong but I don't understand quantitatively what the gear ratio numerator values represent. From gear 1 to 2, the car is set to shift at 81km/h but in actuality, the car had to shift earlier because it hit the 9606rpm it wants to shift at. As I mentioned, the rpm drops to the peak power rpm 6800 and each shift after that drops me 200-300 rpm higher than the last one. I know you're supposed to shorten the gears if you notice that you aren't getting sufficient power between shifts but how can I take the numbers to calculate exactly what rpm to shift at?
So that's a much more scientific approach than I've taken to gearing. Typically when I'm tuning I'll look at two different things when determining my gear ratios. 1) Am I having to shift during cornering? If so, I'll open or close the gear ratios depending on how the car feels on the track, keeping in mind in a lower car you typically get more tire spin with power applied. And 2) am I getting near the top of the rev band on the straights. Typically this usually involved me extending the top two gears slightly to make sure I'm not hitting the limiter on straights. There is definitely an approach where you could maximize the gearing based on optimal bands, but I find that between trying to shirt shift for fuel economy on endurance races, and the two situations I mentioned above, I'm almost never shifting in the "perfect" spot. I think it's more about having the gearing match the track you're running to ensure you can maximize your time.
Watkins Glen long course, pp600 club cup plus... Ive run integras, ford focus, the citron electric hyper car... Best ive done is 10th and i keep noticing every other car has more take off and torque, it feels like i need more gears than 5. Like at least 8 but ive got to be doing something wrong. Any ideas?
The tuning? It mostly applies when your making a car build for a specific track. On any BoP races it doesn’t help at all. But I use tuning anytime I play, I just like adjusting the cars and making specific builds for certain races.
Oh man I would, I don't drag though! I'd imagine as loose on the differential as you could, and raising the back end slightly would help. Weight shifted back as well.
try turning final all way down. then set top speed all way up. tune 1st gear to no wheel spin from full power brake stand. (don't worry about other gears yet) When u have 1st sorted bring the rest down to even gaps. then go through and maximize your rpms, power band and top speed for acceleration and speed.
