Gears

Can you please show the equations 4.85 and 4.89.

4.89 doesn't make sence for me...

Here is everything

regards

Yohann

Ah OK, they just assume that the module for the constant gear and the single gear stages in the gearbox are the same. Which is not correct. You can have several different modules in a gearbox only mating gears need of course the same module.

Anyway, by assuming the module is the same, they came up with all the equations. In the table 4.7 3 different ratios for the constant stage are considered (1,25 1,6 and 2). The rest of the calculations comes from the equations.

It's one possible way, but it's not my favourite.

Hi alex,

I understand what you mean about the modules now. That shouldn't be correct if they have just assumed a constant module. However i still dont understand the process that they have taken.

If i give you some ratios as an example could you please show me what to do (show your working out in writting for a couple of gears so i will understand the method). As mentioned above you said this was not your favorite method to use. If your way is better can you show me i am all ears to learning if that is possible.

Final_Gear_Ratios =

• NL :       14.3783
• Gear 2:   10.0679
• Gear 3:  7.0497
• Gear 4:   4.9364
• Nh:    3.4565

Regards

Yohann

The calculation is really easy, it is just not completly explained.

• The ratio for (e.g.) gear 1 is defined to be 4.
• The ratio of 4 is split into the constant gear N2/N1 and the gearpair NG1/NP1. N2/N2 is assumed to be 1,25.
• The sums N1+N2 and NG1+NP1 hast to be equal (module is same)
• Calculate multiple values of 2,25 and 4,2, since the teeth numbers are integers. The first common multiple value for both 2,25 and 4,2 is the sum of the teeth N.
• Therefore NP1 is 15 and NG1 63-15=48
• N1 is 28 and N2 is 63-28=35

If you do the same calculation for N2/N1=1,6 the sum N will be 91.

The problem is that you have to guess this constant ratio correctly otherwise the will be no common integer for the ratios.

Hi alex,

I have read through what you have sent me, and I have a few more questions as I am still confused.

1. If we know the gear ratio to be “4” then how do, we know to split it up into (1.25*3.2) there could be infinite combinations of numbers that their product is 4 (the gear ratio).
2. Where do we get the numbers “2.25” and “4.2” are they the same for every gear or should they change. Where is the relationship between those numbers and the “1.25” and “3.2”?
3. Even if we do this calculation, we will need to ensure that each set of gears can fit on a certain distance between centres of shafts. I don’t see how we can ensure that they will fit as a constant between the shafts is not mentioned anywhere.
4. Since the module of each gear is different in the transmission how do we know which gear should have what module
5. Table 4.5 and 4.6 show what the max and min of NP and NG should be to prevent interference. How does that relate to the formulas and the design?

I don’t want to seem like an idiot, but I am just very confused all I can see are numbers being plucked out of thin air and somehow, they are in relation to each other. How can that be possible?

I from my calculator I have come up with the gear ratios mentioned above post. I have tried to relate it to that but i have gotten nowhere. I am just very confused.

regards

Yohann

Hi,

1. Yes, there is an infinite amount of combinations. You have to distribute the ratio in a way you are happy with, maybe looking at gear 1 and defining the ratio not to be higher than 3. Therefore the constant will be min. 1,33. Or some other definition like minimum constant of 1,5.
2. 2,25 and 4,2 are the sums N. Since N2/N1=1,25, N2=1,25·N1 and the sum is N1+N2=N1+1,25N1 which is 2,25N1. The same for NP1 and NG1. Just 1,25+1 and 3,2+1.
3. It's insured in this calculation as we assume to have the same module. The equation 4.84 and 4.85 are considering the center distance.
4. The module is defined by the fine gear geometry and gear strength calculations. It is influenced by center distance, gear speeds, torque, profile shifting, no of teeth, aimed safeties for tooth root and tooth flank etc. Usually software is used to do this calculations and try out diverse modules.
5. The tables may be checked after calculating the no. of teeth. The smallest pinion is the 3rd calculation with 11/22 which is still OK for helical gears with beta=25°

In your calculations the overall ratios are shown. As you are working with a standard drivetrain layout with a layshaft 5-speed gearbox your 5th gears ratio inside the gearbox i5=1 (direct gear). Therefore you differential ratio iDiff=3,4565. Divide all your ratios by the diff ratio to recieve the pure gearbox ratios. After it you can distribute the ratios between constant and the rest.

Hi alex,

I finally understand what is going on. You were right the math is very easy however when it wasn't explained properly it just seemed so confusing.

When you talk about the final drive ratio the formula is the (Tire circumference/drive ratio) however as you mention correctly there the last ratio should be 1 as it will be a direct drive however usually aren’t there overdrive gears. Therefore the 2nd last gear should have a ratio of 1 and the last gear (overdrive gear) should be >1. To calculate the overdrive gear is it a completely new formula?

In my calculator I have worked out the highest ratio based of the highest ratio based of the max torque of the engine therefore it won’t be 1. What should I do because I would like to make the calculator complete and add at least one or more overdrive ratios?

regards

Yohann

Hi Yohann,

Here I have to correct some facts from your message. First of all you don't need to look at the individual ratios of other gearboxes and try to achieve the same - You have done your own calculations and they aren't wrong. 14,37 as highest ratio and 3,45 for the 5th gear lead to an overall ratio of ~4,16. It's OK for a racecar.

Further when you design a gearbox you don't care about overdrive - there is nothing like overdrive, it's only a marketing term! They call it overdrive when the gearbox ratio is lower then 1 and the gearbox output turn faster than the input. But the car needs the overall ratio which includes the differential ratio (bevel gear stage, i=3,45 in your case). So overdrive or not, the overall ratio will be greater than 1 and therefore no overdrive.

You have your ratios and can now calculate the gear pairs. No more information is missing!

Hi alex,

I have made another calculator in excel to calulate the K value for converting the gear ratios into number of teeth.

However I just wanted to know should there be a limit on the how high the value of K can go?

regards

Yohann

You're still using the overall ratios - now you calculate gearbox ratios which don't contain the differential ratio.

Notice that you only distribute the ratio between the constant and a certain gear only once. Then you know the ratio of the constant and can easily get the ratios for the other 3 gear pairs.

K will go up as high as necessary but needs to stop if the sum of the teeth gets too big.

Hi alex,

How do I calculate the rest if I have the constant? I thought I would have to do simultaneous equations but I will have 4 different unknowns so that is clearly not correct. What am I missing out?

regards

Yohann

Hi Yohann,

I tried to use this method of calculation myself and I must say... It's pretty shitty :D It was nice to learn a new way of calculation but just forget about it.

Remember I wrote about defining the center distance and than calculating the reference diameters? That's a better way.

• Define a center distance
• Asumme a ratio für the constant
• Calculate the 4 other gear pairs (ref. Ø)
• With the calculated ref. Ø you can achieve the number of teeth by defining a module. Of course there will be a small diviation to your "perfect" defined ratios.
• Compare the calculated ratios to the target ratios and decide whether they are okay. If not, choose other no. of teeth.

Which size of tyres do you use? Assuming a circumference of 2,223m and a max. engine speed of 8000rpm your car runs nearly 300km/h.

Hi Alex,

The book that showed me the previous confusing method also mentioned about a saw tooth diagram. In the previous sections however it was not explained at all it only showed me the exact same graph with no calculations next to it.

Therefore could you explain it like the previous questions. I assume you have worked it out since you have and output graph. So just show me for a few gears and then I should understand the method. You will have to explain what is happening since I will have nothing to reference it to or read.

From what you have said I should work out the centre distance then the ref diameter but the ref diameter depends on number of teeth and module?

The issue I have noticed is that I still haven't taken gradability into account. Is that an issue? Since I will only have taken flat roads into account. Also won't that mean all the calculated values that I have done already be wrong if the method is changing or not?

Regards

Yohann

The saw tooth diagram is nothing special and there is not much to explain. You know all your overall ratios, you know the size of the tyre => You can calculate the car speeds in every gear depending from the engine speed. A geometrical stepped gearbox shows in the saw tooth diagram, that the engine speed alsway drops to the same value when shifting. That's it.

Does the ref. Ø depend on module and number of teeth or can you choose module and no. of teeth depending on the ref. Ø? ;-)

If gradability is important for your vehicle you should recalculate the ratios. You know that you define the 1st and the last speed ratio, when starting gearbox calculations. Here you can take account of several factors. Top speed, power reserve at top speed, fuel saving on high way, acceleration, idler speed, gradability etc... You must define the factors which are necessary for the vehicle.

Hi alex,

I have now managed to calulate everything and produce the relevent graphs and number of teeth.

I am now moving on to calculating the gear geometry and i have come across a problem. When calculating the module for the gears since using DIN standerd I am required to have the module pre defined. so i thought since i know the distance between the gears roughly i would use the equation

A = M(Z1+Z2)/2

M = (A*2)/(Z1+Z2)

and just rearange it so that i could work out the module for each set of gears. As mentioned before only the meshing gears need to have the same module but when i use this method i have every gear having the same module. Is this an issue since when using KHGears there are two diffrent sets of formulas if the there is a profile shift therefore having the same module for each gear would be an issue?

Or can i still use the profile shifted equations but have the values for the profile shift for both gears to be Zero. Is that allowed or not.

regards

Yohann

Yohann Panthakee What is the name of the book ? And its author please.