By Bruce Johnson
(Doesgo on HPISF & former Slap Ma Fro Owner)
This is a very long post, obviously, but when I try to explain something I tend to get wordy! It's not as complex as it appears, I just wanted to make things very clear.
Getting all the tooth counts and keeping them in order is the tough part, the rest is just simple math and it's all about multiplying gear ratios (not adding them).
First, let's use a buggy as a simple example (no transmission) that has the following for gears: 13-tooth bell, 46-tooth spur, 12-tooth pinion, and a 43-tooth ring gear.
Start at the engine and work your way down the drivetrain. Since the bell has fewer teeth than the spur, the engine will spin a few times for every one rotation of the spur, right? How many? Divide the tooth counts: 13 / 46 = 0.283. That means the spur spins 0.283 times for every one rotation of the clutch bell.
Do the same thing all the way through the drivetrain. Take the tooth count of the drive gear and divide it by the tooth count of the driven gear. In our simple example:
Bell (13T) / Spur (46T) = 0.283
Pinion (12T) / Ring Gear (43T) = 0.279
That's it for our buggy example. Now multiply the two ratios together and the result is the number of times the differential (and thus, the wheels) will rotate for every engine rotation:
0.283 x 0.279 = 0.079
Okay, so now we know the wheels do 0.079 rotations per engine rotation. But how far is that down the road? Well, what's the circumference of the tire? Simple: wrap a flexible measuring tape around the tire tread. Math: measure the tire's height (diameter) and multiply by pi (3.14159). If this buggy's tire is 4" in diameter, the circumference is 4.00 x 3.14159 = 12.566". So every time the tire goes around one time, the buggy moves 12.566 inches. Back to the engine revolution...
Since the tire rotates 0.079 times for every engine rev and each FULL tire rotation causes the vehicle to travel 12.566 inches, just multiply those two numbers together to find out how far the vehicle travels for every engine rev: 0.079 x 12.566 = 0.993". For every engine rev, the buggy will move just under one inch.
======================================
How to apply this to the Savage? Same thing, just more numbers involved thanks to the transmission. All you have to do is write down the number of teeth on each gear and note which gears drive (or are driven by) other gears. In the buggy example there were two gear ratios to consider because the spur is connected to the pinion by a dogbone. But if that dogbone disappeared and the pinion was driven directly by the spur's teeth, you'd have to add that ratio to the equation. In other words, when there's a driveshaft between two gears, the drive ratio between the gears is 1:1, and as you know, anything multiplied by one doesn't change anything, so we just ignore it.
I don't know the gearing layout of the Savage transmission, but I'll throw out some numbers so we can walk through it.
Bell: 13T
Spur: 47T
Internal #1: 16T
Internal #2: 20T
Internal #3: 24T
Pinion: 13T
Ring: 43T
But what connects to what via what method? This part is crucial to obtaining the correct numbers!
Bell/spur connect via gear teeth. Spur/Internal #1 connect via shaft (no ratio needed). I#1/I#2 via teeth. I#2/I#3 via teeth. I#3/pinion via shaft (no ratio needed). pinion/ring via teeth. Thus the equation:
(13 / 47) x (16 / 20) x (20 / 24) x (13 / 43) = 0.055, which is 0.055 tire rotations for every engine rotation. Assuming a 6" diameter tire gives us a circumference of (6" x 3.14159) 18.850". Take that times the 0.055 to get the total distance travelled per engine rotation: 0.055 x 18.850 = 1.037". For every engine rev, the Savage will move just over one inch.
====================================
Remember, those Savage numbers are theoretical. You need to plug the tooth counts for your vehicle, and only use the transmission gears involved. If you want the answer to related to when the truck is in third gear, only use the internal tranny gears that are involved when using 3rd gear, and determine which are connected via their teeth and which are connected via a common shaft. Also, if there's an idler gear involved (a gear that's only there for further gear reduction or to take up space, such as the middle gear on a three-gear setup like that in most stadium trucks), you need to include it in the equation.
Also, don't forget to measure your tires and plug that number in as well. Feel free to post all your info here and I can help with the calculation if you wish, or just PM me.
Oh, one more thing: tire growth! You've seen tires ballooning when a truck "diffs out," correct? Well, the faster an RC vehicle goes, the taller the tires get. It's not very noticeable until usually over 30mph or so (on trucks), but if you're into top speed, it makes a difference. The circumference of the tire increases, making the vehicle go farther per engine revolution. Here's a shot of Masher 2000 tire growth on the back of my RC10GT...even the front tires were ballooning slightly!
(Doesgo on HPISF & former Slap Ma Fro Owner)
This is a very long post, obviously, but when I try to explain something I tend to get wordy! It's not as complex as it appears, I just wanted to make things very clear.
Getting all the tooth counts and keeping them in order is the tough part, the rest is just simple math and it's all about multiplying gear ratios (not adding them).
First, let's use a buggy as a simple example (no transmission) that has the following for gears: 13-tooth bell, 46-tooth spur, 12-tooth pinion, and a 43-tooth ring gear.
Start at the engine and work your way down the drivetrain. Since the bell has fewer teeth than the spur, the engine will spin a few times for every one rotation of the spur, right? How many? Divide the tooth counts: 13 / 46 = 0.283. That means the spur spins 0.283 times for every one rotation of the clutch bell.
Do the same thing all the way through the drivetrain. Take the tooth count of the drive gear and divide it by the tooth count of the driven gear. In our simple example:
Bell (13T) / Spur (46T) = 0.283
Pinion (12T) / Ring Gear (43T) = 0.279
That's it for our buggy example. Now multiply the two ratios together and the result is the number of times the differential (and thus, the wheels) will rotate for every engine rotation:
0.283 x 0.279 = 0.079
Okay, so now we know the wheels do 0.079 rotations per engine rotation. But how far is that down the road? Well, what's the circumference of the tire? Simple: wrap a flexible measuring tape around the tire tread. Math: measure the tire's height (diameter) and multiply by pi (3.14159). If this buggy's tire is 4" in diameter, the circumference is 4.00 x 3.14159 = 12.566". So every time the tire goes around one time, the buggy moves 12.566 inches. Back to the engine revolution...
Since the tire rotates 0.079 times for every engine rev and each FULL tire rotation causes the vehicle to travel 12.566 inches, just multiply those two numbers together to find out how far the vehicle travels for every engine rev: 0.079 x 12.566 = 0.993". For every engine rev, the buggy will move just under one inch.
======================================
How to apply this to the Savage? Same thing, just more numbers involved thanks to the transmission. All you have to do is write down the number of teeth on each gear and note which gears drive (or are driven by) other gears. In the buggy example there were two gear ratios to consider because the spur is connected to the pinion by a dogbone. But if that dogbone disappeared and the pinion was driven directly by the spur's teeth, you'd have to add that ratio to the equation. In other words, when there's a driveshaft between two gears, the drive ratio between the gears is 1:1, and as you know, anything multiplied by one doesn't change anything, so we just ignore it.
I don't know the gearing layout of the Savage transmission, but I'll throw out some numbers so we can walk through it.
Bell: 13T
Spur: 47T
Internal #1: 16T
Internal #2: 20T
Internal #3: 24T
Pinion: 13T
Ring: 43T
But what connects to what via what method? This part is crucial to obtaining the correct numbers!
Bell/spur connect via gear teeth. Spur/Internal #1 connect via shaft (no ratio needed). I#1/I#2 via teeth. I#2/I#3 via teeth. I#3/pinion via shaft (no ratio needed). pinion/ring via teeth. Thus the equation:
(13 / 47) x (16 / 20) x (20 / 24) x (13 / 43) = 0.055, which is 0.055 tire rotations for every engine rotation. Assuming a 6" diameter tire gives us a circumference of (6" x 3.14159) 18.850". Take that times the 0.055 to get the total distance travelled per engine rotation: 0.055 x 18.850 = 1.037". For every engine rev, the Savage will move just over one inch.
====================================
Remember, those Savage numbers are theoretical. You need to plug the tooth counts for your vehicle, and only use the transmission gears involved. If you want the answer to related to when the truck is in third gear, only use the internal tranny gears that are involved when using 3rd gear, and determine which are connected via their teeth and which are connected via a common shaft. Also, if there's an idler gear involved (a gear that's only there for further gear reduction or to take up space, such as the middle gear on a three-gear setup like that in most stadium trucks), you need to include it in the equation.
Also, don't forget to measure your tires and plug that number in as well. Feel free to post all your info here and I can help with the calculation if you wish, or just PM me.
Oh, one more thing: tire growth! You've seen tires ballooning when a truck "diffs out," correct? Well, the faster an RC vehicle goes, the taller the tires get. It's not very noticeable until usually over 30mph or so (on trucks), but if you're into top speed, it makes a difference. The circumference of the tire increases, making the vehicle go farther per engine revolution. Here's a shot of Masher 2000 tire growth on the back of my RC10GT...even the front tires were ballooning slightly!
Last edited by a moderator:
