What is the entire gear reduction ratio of a vehicle?

[FrancoisL]

TL;DR

What represents the entire reduction ratio of a vehicule, and how is it related to other value that act in a vehicule gear system (Torque, RPM,gear ratio...)?

As I was looking in the caracteristics of an old tank, I found a number that I could not get anything of on the web. It is called "entire reduction ratio" and was presented just under the gearbox caracteristics. His value is 1:13.4. My question is : is it useful? Does this value have relation whit other quantities in a gear vehicule system (For example, Torque, RPM of motor or the wheels, gear ratio, Final drive ratio)?

 

[Lnewqban]

Welcome, FrancoisL!

Please, see:
https://en.m.wikipedia.org/wiki/Gear_train

The phrase is applicable to one gear box, to a sequence of two or more gear boxes, or to the sequence of gear box, differential and wheels of any vehicle.

In most cases, the rotational speed of an engine is too high compared with the speed at which the wheels spin.
At the same time, the torque demanded by the terrain at the wheels is huge compared with the torque that an engine can deliver.

Because of the above, gear trains that reduce the rotational speed (as increase the torque) are necesary in most vehicles.
The ratio of your tank could mean that one output shaft rotates 10 turns each time the input shaft rotates 134 turns.

 

[FrancoisL]

Hi, thanks a lot, I'm sure I will love Physics Forum

if I understand well, this gear is there to reduce the speed of the engine and increase the torque (because it is needed), exactly as the final drive and gear box ratios do. Do you know how I could include it in the calcul from speed at the engine (rpm) to speed at the wheel (km/h, I have the Wheel Radius)? If I try to put it in a logical place (dividing the Speed of the engine), my speed at the wheels is too slow.

 

[jack action]

No, the "entire reduction ratio" is not related to a single gear set. It most likely is an engine shaft that goes into a gearbox with a gear ratio of 1:3.35 and the gearbox goes into the final drive with a gear ratio 1:4. And that final drive output shaft is connected to the wheels. The entire reduction ratio (from engine to wheels) is 1:13.4, because 3.35 X 4 = 13.4.

 

[FrancoisL]

No, the "entire reduction ratio" is not related to a single gear set. It most likely is an engine shaft that goes into a gearbox with a gear ratio of 1:3.35 and the gearbox goes into the final drive with a gear ratio 1:4. And that final drive output shaft is connected to the wheels. The entire reduction ratio (from engine to wheels) is 1:13.4, because 3.35 X 4 = 13.4.

I'm not totally sure to understand :-(, sorry.
Here's the gear ratios of the vehicule (attached file, first column)
Final Drive : 8.4:1 or 1:8.4, I have two contradictory source.
How do I get to 1:13.4?
If it is just a multiplication of data I already know, I won't use it in calculus, but can you confirm with the data I gave this is that?
Thanks a lot for details anyway! I know now it is a reduction ratio grouping multiple gear (a whole gear system) and no just one gear

.

 

[jack action]

Looking at your data, it seems to be the range for the gear ratios of the gearbox. The first gear is 9.21 and the last one is 0.69, thus 9.21/0.69 $\approx$ 13.4. This means that you can cover the same speed/torque range than with a transmission with first gear 13.4 and last gear 1 (you would only need a different final drive).

 

[FrancoisL]

That is certainly the explanation. Thanks a lot for your help!

 

[Lnewqban]

Hi, thanks a lot, I'm sure I will love Physics Forum

if I understand well, this gear is there to reduce the speed of the engine and increase the torque (because it is needed), exactly as the final drive and gear box ratios do. Do you know how I could include it in the calcul from speed at the engine (rpm) to speed at the wheel (km/h, I have the Wheel Radius)? If I try to put it in a logical place (dividing the Speed of the engine), my speed at the wheels is too slow.

You can do the calculation with precision, but only if you know all the links in the chain of reductions that connects the engine's shaft with the wheel's shaft.

Please, see:
https://courses.lumenlearning.com/physics/chapter/6-1-rotation-angle-and-angular-velocity/

I am not familiar with the internal mechanisms of tanks, butI assume they need a lot of torque supplied at the wheels just to cope with the obstacles and irregularities of the terrain, as well as with the massive inertia.

Do you know the meaning of that "Jump" in the picture you have posted above?

 

[sophiecentaur]

His value is 1:13.4. My question is : is it useful? Does this value have relation whit other quantities in a gear vehicule system (For example, Torque, RPM of motor or the wheels, gear ratio, Final drive ratio)?

To know what the tank can actually do - speed, pushing force etc - you have to know the radius of the drive wheels, as you have said. Also, it is necessary to know about the engine characteristics for Power and Torque and many engines (probably even more so for ancient designs).

You say you have attempted to include the wheel radius but obtained a wrong (?) answer . In what way was the answer wrong? Does the transmission system of a tank involve a differential as in a normal road vehicle? That will always include some sort of 90 degree change of direction of the shaft* and there is always (?) an additional ('back axle') gear reduction. Have you included this?
*Crown wheel and pinion system.

 

[FrancoisL]

Hi,

No, I don't know the meaning of jump, I was probably going to be my next question. There is no other information on this jump column on the sheet paper (I'll share all the data I have as attached file), so if anyone knows about it, feel free to help us.

Hi, SophieCentaur, I attempted some calculs, from engine RPM to wheels speed, and every time my answer was about 33.8% higher than the real (the one seen on the field) value. It was never more than 34.5% higher and never less than 32.5% (rougthly). I had 16 RPM engine and the real speed associated to it (the one seen in the table in my previous message). My calculs where done like this :

(rpmEngine*WheelCircumference)/(gearRatio*FinalDriveRatio)

Whit some factor to adapt the physics units. The wheel radius is 430mm (860 of diameter). As you can see, I haven't include a back axle ratio.

I also have a graph with the Pme of the engine (RalDhFY), allowing to calculate the torque (4 stroke engine, 23095cm^3 of engine dispacement), which revealed to be 8% lower than the real value (but I had only one value unfortunately, it doesnot mean that much). I included it aswell, it might help. The 100% of the bottom correspond to 3000rpm (and so 70% is 2100rpm), Drehzahl is rpm and Dremoment is torque (but I did not tried to read that graph, the % unit for Torque scrared me, so if you know how to read it...).

So, the tank has a Maybach double differential, it is a double controlled differential and I definitely don't know exactly to which point it is similar to a car differential.
https://en.wikipedia.org/wiki/Tank_steering_systems (scroll down to Maybach double differential).

I have a last interesting graph if you want to see (last doc, diagram 2).

Thanks a lot for the interest on my calculs, as I was not sure of them, some advise from expert is welcome in my case.

 

[jack action]

No, I don't know the meaning of jump, I was probably going to be my next question.

Same thing as "entire reduction ratio", but for one gear to the next.

$$\frac{9.21}{4.56} = 2.02$$
$$\frac{4.56}{2.87} = 1.59$$
$$\frac{2.87}{1.83} = 1.56$$
$$\frac{1.83}{1.27} = 1.45$$
$$\frac{1.27}{0.90} = 1.41$$
$$\frac{0.90}{0.69} = 1.31$$
$$2.02×1.59×1.56×1.45×1.41×1.31 = 13.4$$

 

[FrancoisL]

Same thing as "entire reduction ratio", but for one gear to the next.

$$\frac{9.21}{4.56} = 2.02$$
$$\frac{4.56}{2.87} = 1.59$$
$$\frac{2.87}{1.83} = 1.56$$
$$\frac{1.83}{1.27} = 1.45$$
$$\frac{1.27}{0.90} = 1.41$$
$$\frac{0.90}{0.69} = 1.31$$
$$2.02×1.59×1.56×1.45×1.41×1.31 = 13.4$$

Nice!

 

[Lnewqban]

You should use radians/second rather than rpm's.
Please, check the link I posted above.

 

[FrancoisL]

Yes, you link is good, it helped me to understand the formula I used, but I don't understand why I should use radian rather than Rpm (I mean, all my values are rpm, why bother convert them if my formulas used rpm). I'm not a physicist, so I don't really know, is the result more accurate using radians?

 

[jbriggs444]

Yes, you link is good, it helped me to understand the formula I used, but I don't understand why I should use radian rather than Rpm (I mean, all my values are rpm, why bother convert them if my formulas used rpm). I'm not a physicist, so I don't really know, is the result more accurate using radians?

Using radians per second gets rid of a mildly annoying conversion factor if you try to use torque and rotation rate to compute power. Or when you try to use rotation rate and wheel radius to compute linear speed.

Us math types get used to radians and have trouble imagining any other way to do business.

Us Americans saddled with incoherent unit systems (horspower, pounds force, pounds mass, rpm, feet per second, acres) get used to conversion factors. One more is no big deal.

[I have no problem taking both sides of any given argument]

 

[sophiecentaur]

I don't understand why I should use radian rather than Rpm (I mean, all my values are rpm, why bother convert them if my formulas used rpm).

I agree with you. You are after a specific answer so you can just work out the length traveled by one rev of the drive wheel and that would give you a speed. (There may be some compensation needed because the radius where the teeth of the drive apply the most force on the track segments.)
The only problem that I can see is the Double differential arrangement which has different final drive ratios for front and back. Front and rear axles would go at the same revs (of course) but the share of the drive torque wouldn't be equal. This doesn't surprise me because my Landrover Freelander (a very humble off roader) has a similar arrangement with a (shockingly overpriced) system for sharing the front and rear drive torque on what looks to be a variable basis.
For your problem I'm not sure what the effective final drive ratio would be - probably some sort of mean between the two but not the arithmetic mean. Some clever clogs on PF will probably know the answer to this so I will rest my brain on this one. But it may explain your odd results.

PS Don't those documents together already tell you quite enough information about those big boys?

 

[FrancoisL]

Yes, your are certainly right, these documents are a good representation of the system making this tank work. I have a few more sources were I can try to find the answer I was looking for, but I think I have enought information. Thanks a lot for you research, you all helped me a lot with this.