If the front & rear weights were the same then wouldn't the rear see more effective power due to weight transfer? Also the front shafts were smaller on my 203 equipted trucks as well as my 205/208 trucks.
CK5
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Does anybody know how the power is split up between the front and rear?
- Thread starter PetaKane
- Start date
shaft torque is simply the wrong term to use in this discussion as was stated by Triaged.
torque requires a resistance, no?
if the rear axle is suspended and the front is firmly planted how much torque is being transmitted to each axle?
I would say near 100% to front and the rear is freewheeling.
either way the shafts are solidly linked in parttime TCs available in these trucks
torque requires a resistance, no?
if the rear axle is suspended and the front is firmly planted how much torque is being transmitted to each axle?
I would say near 100% to front and the rear is freewheeling.
either way the shafts are solidly linked in parttime TCs available in these trucks
I'm not sure, that is why I used "power" instead of "torque" in my post. From what I read in the dictionary his decription is wrong. It only states that torque is a rotational force. It references the axis of the spinning unit but nothing on where the outer edge would be, nor any load placed on the shaft. I may need to look in a different (technical) book though. </font><blockquote><font class="small">In reply to:</font><hr />
either way the shafts are solidly linked in parttime TCs available in these trucks
[/ QUOTE ] Agreed.
either way the shafts are solidly linked in parttime TCs available in these trucks
[/ QUOTE ] Agreed.
Thumper
1/2 ton status
</font><blockquote><font class="small">In reply to:</font><hr />
front would see more torque in my opinion due to the weight over the front,
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But isnt that counteracted as soon as the truck starts transferring weight to the rear when you start off? I would think that much and more are onto the rear very quickly when you punch it. If that happens, it may seem like less is sent to the front, but its still the same. You see these guys on the rocks, sitting on a rock hillside, all 4 tires slowly turning at exactly the same speed. This wouldnt be possible if the case split it anyother way than 50/50. I think by the sounds of everyone, the power isnt split any other way but 50/50 front to rear. The only variable is the center of gravity as it moves back and forth depending on the attitude of the truck. That answers my question... and I think that was the original one. The way I understood it anyways.
Mike
front would see more torque in my opinion due to the weight over the front,
[/ QUOTE ]
But isnt that counteracted as soon as the truck starts transferring weight to the rear when you start off? I would think that much and more are onto the rear very quickly when you punch it. If that happens, it may seem like less is sent to the front, but its still the same. You see these guys on the rocks, sitting on a rock hillside, all 4 tires slowly turning at exactly the same speed. This wouldnt be possible if the case split it anyother way than 50/50. I think by the sounds of everyone, the power isnt split any other way but 50/50 front to rear. The only variable is the center of gravity as it moves back and forth depending on the attitude of the truck. That answers my question... and I think that was the original one. The way I understood it anyways.
Mike
Chris_T
1/2 ton status
Just a take on the perception thing - an open diff can make it look like a tire is going slower than the rest. If your passenger side tire is spinning freely at the same speed as the rear axle because of lack of traction, your driver's side front can still be turning, albeit slower. It's not perception - it really is slower, but the tire on the other side is movingthe same as the rears
Triaged
1/2 ton status
</font><blockquote><font class="small">In reply to:</font><hr />
Or maybe its just physics. The weight transfer of the vehicle when the power is being applied makes it nearly impossible to break even a small front shaft because the front tires lose so much ground pressure? So, the rear shaft is made to be able to handle anything the motor can throw at it, and the front is designed so that it can handle as much as it needs to?
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You got it right there.
Most people drive forward up hills or tow while going forward.
This causes alot of weight to transfer to the rear which increases the traction it has and also the ammount of torque before the wheel slips.
So the front driveshaft does not need to be as thick because:
1) It is not used as much as the rear (given a part time t-case)
2) Because of clearance issues (on non-lifted trucks)
3) Most of the time you go forward and transfer weight off the front end
4) Other parts would break in the front before the d-shaft anyway.
And to state my point in case you missed it:
A locked t-case will almost NEVER put out a 50:50 torque split
A 203 will almost ALWAYS put a 50:50 torque split (when not locked).
Dicussion closed.
Or maybe its just physics. The weight transfer of the vehicle when the power is being applied makes it nearly impossible to break even a small front shaft because the front tires lose so much ground pressure? So, the rear shaft is made to be able to handle anything the motor can throw at it, and the front is designed so that it can handle as much as it needs to?
[/ QUOTE ]
You got it right there.
Most people drive forward up hills or tow while going forward.
This causes alot of weight to transfer to the rear which increases the traction it has and also the ammount of torque before the wheel slips.
So the front driveshaft does not need to be as thick because:
1) It is not used as much as the rear (given a part time t-case)
2) Because of clearance issues (on non-lifted trucks)
3) Most of the time you go forward and transfer weight off the front end
4) Other parts would break in the front before the d-shaft anyway.
And to state my point in case you missed it:
A locked t-case will almost NEVER put out a 50:50 torque split
A 203 will almost ALWAYS put a 50:50 torque split (when not locked).
Dicussion closed.
Your description of torque contradicts what my dictionary states. It mentions nothing about friction or load. Please tell us (me) where where you found your info, ie text book ,net , tech documents, ect. Trying to follow but your last two statments seem wrong.
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Trying to follow but your last two statments seem wrong.
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agreed, seems never and always should swap places /forums/images/graemlins/confused.gif
Trying to follow but your last two statments seem wrong.
[/ QUOTE ]
agreed, seems never and always should swap places /forums/images/graemlins/confused.gif
wrathORC
1/2 ton status
Nah, not for part time cases. In part time cases the front and rear driveshafts are tied together. There is no differential, viscous clutch, or wet clutch to bias torque from front to rear.
However, which tire it goes to is a different story with open differentials. You're always going to break the front left and right rear loose if all four pieces of rubber have the same amount of friction with the ground. This is because of the way a differential works and the direction the driveshaft rotates. I'm not quite sure how to go about explaining it since I'm not a Physics major. I'm pretty sure I'd just end up confusing both of us if I did try.
However, which tire it goes to is a different story with open differentials. You're always going to break the front left and right rear loose if all four pieces of rubber have the same amount of friction with the ground. This is because of the way a differential works and the direction the driveshaft rotates. I'm not quite sure how to go about explaining it since I'm not a Physics major. I'm pretty sure I'd just end up confusing both of us if I did try.
R72K5
Banned
also keep in mind that in at least the 72 older years the front axles on 3/4 4x4 trucks got 4.55
whereas the eaton rears had 4.57 in them and dana 60s had 4.56 and there was no 4.55 rear ratio at all.
whereas the eaton rears had 4.57 in them and dana 60s had 4.56 and there was no 4.55 rear ratio at all.
If there is I don't know how accurate it could be.
In part time trucks, the front and rear get 50/50 (period). One front tire may seem to go slower than one on the rear axles, and it can and WILL happen especially if one or both axles has an open diff or limited slip. On a rear axle equipped with and open diff, with enough driveshaft input, if one rear tire is hardly spinning, then the other rear wheel will spin faster (similar to turning with an open diff, one wheel WILL spin faster, but both wheels put together are turning in a direct relation to the input speed of the driveshaft). With an open diff in the front, one front wheel may hardly spin but the other goes nuts, and the back wheels (say if equal traction or have a locker) with both spin at the same rate. The rears will turn slower than the fast front wheel, but faster than the slow front wheel. This is ALL assuming there is loss of traction at one or more wheels, BUT both driveshafts are turning the same speed, and the NET result of Wheelspin are equal in the front as it is in the back.
With a non converted 203 in non loc mode, there is no formula such as 50/50, 60/40, etc. It all depends on weight transfer and traction. It is variable just like the open diff in a rear axle. Sometimes the rear gets all the power, sometimes the front gets all the power, and other times it is anywhere in between depending on where the traction is (or isn't), all in an effort to gain more traction without the side effect of driveline bind associated with traditional 50/50 split part time cases.
On some AWD cars with "fluid couplers", they advertise 60/40 splits or X/Y splits, but I think it is marketing mumbo-jumbo. One axle gets the majority of the traction unless slip is detected and then more power is transferred to the axle that has traction (unless neither do). Instead of an open diff in the transfer case such as the 203, they are closer to a LSD setup which utilizes fluid (or other media). They also use the ABS system to "power brake" and force input power to be distributed accordingly, so traction is not determined just by t-case output and ground conditions, but also from the braking system.
The only reliable number for a split is a true part time case that uses gears or chains to transfer power, and unless something breaks is as close to 50/50 as you can get.
Hope this helps.
In part time trucks, the front and rear get 50/50 (period). One front tire may seem to go slower than one on the rear axles, and it can and WILL happen especially if one or both axles has an open diff or limited slip. On a rear axle equipped with and open diff, with enough driveshaft input, if one rear tire is hardly spinning, then the other rear wheel will spin faster (similar to turning with an open diff, one wheel WILL spin faster, but both wheels put together are turning in a direct relation to the input speed of the driveshaft). With an open diff in the front, one front wheel may hardly spin but the other goes nuts, and the back wheels (say if equal traction or have a locker) with both spin at the same rate. The rears will turn slower than the fast front wheel, but faster than the slow front wheel. This is ALL assuming there is loss of traction at one or more wheels, BUT both driveshafts are turning the same speed, and the NET result of Wheelspin are equal in the front as it is in the back.
With a non converted 203 in non loc mode, there is no formula such as 50/50, 60/40, etc. It all depends on weight transfer and traction. It is variable just like the open diff in a rear axle. Sometimes the rear gets all the power, sometimes the front gets all the power, and other times it is anywhere in between depending on where the traction is (or isn't), all in an effort to gain more traction without the side effect of driveline bind associated with traditional 50/50 split part time cases.
On some AWD cars with "fluid couplers", they advertise 60/40 splits or X/Y splits, but I think it is marketing mumbo-jumbo. One axle gets the majority of the traction unless slip is detected and then more power is transferred to the axle that has traction (unless neither do). Instead of an open diff in the transfer case such as the 203, they are closer to a LSD setup which utilizes fluid (or other media). They also use the ABS system to "power brake" and force input power to be distributed accordingly, so traction is not determined just by t-case output and ground conditions, but also from the braking system.
The only reliable number for a split is a true part time case that uses gears or chains to transfer power, and unless something breaks is as close to 50/50 as you can get.
Hope this helps.
Triaged
1/2 ton status
This is some info about non-locked center diffs.
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TORQUE, RPM, AND POWER DISTRIBUTION IN DIFFERENTIALS
I would like some clarification on the issue of torque distribution between the front and rear axles on 4wd vehicles. I find the matter fairly easy to understand when you have wheels spinning, and a limited-slip differential, but I find it more confusing when I read statements that a vehicle has a permanent torque distribution of, say, 32% front and 68% rear.
To me, torque and revolutions go hand in hand: reduce rpm and you increase torque, as in a ring and pinion. Doesn’t that mean that if you want different torque at the front and rear axles, they have to turn at different speeds?
I know that in vehicles with viscous coupling drive to one axle, one can have a different overall drive ratio at each end, and this is often deliberately employed just to load the system in normal driving, and make it respond quicker to traction loss. But how does a rigid system, with a planetary differential for example, split torque unequally?
When we are dealing with one input torque, from one gear or shaft, and one output torque on a single shaft or other member, the relationship you describe between torque and speed does hold. Neglecting friction, power in equals power out. If rpm is changed, torque must change too, in inverse proportion, for the product of the two (power) to remain constant.
However, when the output power is divided between two shafts by a differential, things change a bit. Total power in still equals total power out (again neglecting friction), but power at each of the two output shafts is not necessarily equal to power at the other shaft. Any non-locking differential maintains a fixed distribution of torque between the two output shafts, while letting their relative speeds vary freely. In a conventional differential, the torque split is 50/50. In a planetary differential with one planetary gearset, the torque split is unequal but still fixed, while the shafts can turn at different speeds.
Usually the differential carrier or planet carrier is driven by a gear, which receives power from another gear driven by the input shaft. At the carrier, the simple inverse relationship between speed and torque applies. Torque at the carrier is input torque times rpm reduction factor. The sum of the output torques equals the carrier torque. The average of the output speeds equals the carrier speed. Power at each individual output shaft can be any value at all. It is even possible to have negative power (retardation) at one output shaft if that shaft is being forced to turn backward (opposite to torque). But the sum of the two power outputs must equal the power input. (That’s the sum of their signed values, not their absolute values.)
It is helpful to think of each spider or planet gear as being similar to a beam, with a load applied at its midpoint, and reaction or support forces at two points equidistant from the load. The load is the drive force applied at the spider or planet gear’s shaft. The reaction forces are the output shaft resistances to vehicle motion, acting at the points of mesh between spider and side gears, or between planet and sun and planet and annulus. Since the spider or planet shaft is always at the gear’s center, the forces at the mesh points are always equal. This is true regardless of the rotational speeds of the various elements.
In a conventional differential, the side gears are equal diameter, so the equal forces at the mesh points act on equal moment arms, and produce equal torques. In a planetary, the annulus is larger than the sun, so the output torque at the annulus is greater than the output torque at the sun. The ratio of the output torques is the ratio of the pitch diameters of the annulus and sun. So the bigger the planet gears are in comparison to the sun, the more unequal the torque split becomes. Usually, the annulus drives the rear axle and the sun drives the front axle.
We can, in fact, regard the conventional differential as a unique version of the planetary, cleverly reconfigured by the use of bevel gears to allow the sun and annulus to be the same size.
All of this determines the torques at the front and rear drive shafts. Usually, the main rpm reduction and torque multiplication (after the transmission) happens at the axle, not at the transfer case. It is
possible to use different ring and pinion ratios at the front and rear axles, and/or different tire sizes front and rear, and further alter the drive force distribution at the tire contact patches. At the axles,
the usual rpm/torque inverse proportionality applies. To get more front torque and less rear by using dissimilar axle ratios, the front drive shaft must turn faster than the rear. That will increase wear at the center diff, rather like traveling a long distance with unequal size tires on an axle. Actually, the least wear at the center diff comes with slightly less torque multiplication at the front axle than at the rear – say a 4.10:1 ring and pinion at the front and a 4.11 at the rear. This is because even on a straight road, the car doesn’t quite go perfectly straight, and in most turns the front wheels will track outside the rears. Consequently, the front wheels travel a few more revolutions per mile more than the rears, even if the effective radii of the tires are equal.
A spool or completely locked differential drives both output shafts at the same rpm, and does not split the torque in any fixed proportion. This is opposite to an open differential, which controls relative torque at the output shafts but not relative speed. With a spool, torque distribution depends on relative resistance at the two output shafts. It is quite possible for one output shaft to have negative resistance (wheel dragging and trying to drive the axle), while the other output shaft has a torque greater than the sum of the two (wheel driving the car plus overcoming drag from the other wheel). The former condition exists on the outside wheel, and the latter on the inside wheel, when making a turn with a spool and no tire stagger.
A partially locking or limited-slip differential is midway between. It allows some difference in speed, but adds torque to the slower output shaft and takes that torque from the faster output shaft.
A viscous coupling transmits torque according to the amount of slippage at the coupling. The faster the input shaft turns relative to the output, the greater the torque at the output shaft. Unlike a gear set, however, the relationship is usually not a simple linear function of the rpm ratio.
Note that none of these alternatives split power equally. No known passive mechanical device does that.
[/ QUOTE ]
</font><blockquote><font class="small">In reply to:</font><hr />
TORQUE, RPM, AND POWER DISTRIBUTION IN DIFFERENTIALS
I would like some clarification on the issue of torque distribution between the front and rear axles on 4wd vehicles. I find the matter fairly easy to understand when you have wheels spinning, and a limited-slip differential, but I find it more confusing when I read statements that a vehicle has a permanent torque distribution of, say, 32% front and 68% rear.
To me, torque and revolutions go hand in hand: reduce rpm and you increase torque, as in a ring and pinion. Doesn’t that mean that if you want different torque at the front and rear axles, they have to turn at different speeds?
I know that in vehicles with viscous coupling drive to one axle, one can have a different overall drive ratio at each end, and this is often deliberately employed just to load the system in normal driving, and make it respond quicker to traction loss. But how does a rigid system, with a planetary differential for example, split torque unequally?
When we are dealing with one input torque, from one gear or shaft, and one output torque on a single shaft or other member, the relationship you describe between torque and speed does hold. Neglecting friction, power in equals power out. If rpm is changed, torque must change too, in inverse proportion, for the product of the two (power) to remain constant.
However, when the output power is divided between two shafts by a differential, things change a bit. Total power in still equals total power out (again neglecting friction), but power at each of the two output shafts is not necessarily equal to power at the other shaft. Any non-locking differential maintains a fixed distribution of torque between the two output shafts, while letting their relative speeds vary freely. In a conventional differential, the torque split is 50/50. In a planetary differential with one planetary gearset, the torque split is unequal but still fixed, while the shafts can turn at different speeds.
Usually the differential carrier or planet carrier is driven by a gear, which receives power from another gear driven by the input shaft. At the carrier, the simple inverse relationship between speed and torque applies. Torque at the carrier is input torque times rpm reduction factor. The sum of the output torques equals the carrier torque. The average of the output speeds equals the carrier speed. Power at each individual output shaft can be any value at all. It is even possible to have negative power (retardation) at one output shaft if that shaft is being forced to turn backward (opposite to torque). But the sum of the two power outputs must equal the power input. (That’s the sum of their signed values, not their absolute values.)
It is helpful to think of each spider or planet gear as being similar to a beam, with a load applied at its midpoint, and reaction or support forces at two points equidistant from the load. The load is the drive force applied at the spider or planet gear’s shaft. The reaction forces are the output shaft resistances to vehicle motion, acting at the points of mesh between spider and side gears, or between planet and sun and planet and annulus. Since the spider or planet shaft is always at the gear’s center, the forces at the mesh points are always equal. This is true regardless of the rotational speeds of the various elements.
In a conventional differential, the side gears are equal diameter, so the equal forces at the mesh points act on equal moment arms, and produce equal torques. In a planetary, the annulus is larger than the sun, so the output torque at the annulus is greater than the output torque at the sun. The ratio of the output torques is the ratio of the pitch diameters of the annulus and sun. So the bigger the planet gears are in comparison to the sun, the more unequal the torque split becomes. Usually, the annulus drives the rear axle and the sun drives the front axle.
We can, in fact, regard the conventional differential as a unique version of the planetary, cleverly reconfigured by the use of bevel gears to allow the sun and annulus to be the same size.
All of this determines the torques at the front and rear drive shafts. Usually, the main rpm reduction and torque multiplication (after the transmission) happens at the axle, not at the transfer case. It is
possible to use different ring and pinion ratios at the front and rear axles, and/or different tire sizes front and rear, and further alter the drive force distribution at the tire contact patches. At the axles,
the usual rpm/torque inverse proportionality applies. To get more front torque and less rear by using dissimilar axle ratios, the front drive shaft must turn faster than the rear. That will increase wear at the center diff, rather like traveling a long distance with unequal size tires on an axle. Actually, the least wear at the center diff comes with slightly less torque multiplication at the front axle than at the rear – say a 4.10:1 ring and pinion at the front and a 4.11 at the rear. This is because even on a straight road, the car doesn’t quite go perfectly straight, and in most turns the front wheels will track outside the rears. Consequently, the front wheels travel a few more revolutions per mile more than the rears, even if the effective radii of the tires are equal.
A spool or completely locked differential drives both output shafts at the same rpm, and does not split the torque in any fixed proportion. This is opposite to an open differential, which controls relative torque at the output shafts but not relative speed. With a spool, torque distribution depends on relative resistance at the two output shafts. It is quite possible for one output shaft to have negative resistance (wheel dragging and trying to drive the axle), while the other output shaft has a torque greater than the sum of the two (wheel driving the car plus overcoming drag from the other wheel). The former condition exists on the outside wheel, and the latter on the inside wheel, when making a turn with a spool and no tire stagger.
A partially locking or limited-slip differential is midway between. It allows some difference in speed, but adds torque to the slower output shaft and takes that torque from the faster output shaft.
A viscous coupling transmits torque according to the amount of slippage at the coupling. The faster the input shaft turns relative to the output, the greater the torque at the output shaft. Unlike a gear set, however, the relationship is usually not a simple linear function of the rpm ratio.
Note that none of these alternatives split power equally. No known passive mechanical device does that.
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ok here is what is getting me hung up. "A foot pound of torque is the twisting force necessary to support a one pound weight on a weightless horizontal bar, one foot from the fulcrum." There is no mention of friction.
here is more "Notice that the torque units contain a distance and a force. To calculate the torque, you just multiply the force by the distance from the center."
I want to see a formula that includes the resistance being figured in.
here is more "Notice that the torque units contain a distance and a force. To calculate the torque, you just multiply the force by the distance from the center."
I want to see a formula that includes the resistance being figured in.
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I want to see a formula that includes the resistance being figured in.
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There is none for calculating torque that I know of. There is a formula that equates resistance and torque, which is along the lines of Torque must be > than the Resistance in order for work to be performed.
In a differenential, torque is neither increased or decreased due to resistance. Torque in equals torque out (assuming no parasitic losses). In a single lever design, resistance never varies the value of torque. In a system where there is 1 torque input, and 2 or more torque outputs, resistance does play a role. It does NOT decrease or increase overall torque (as energy cannot be created nor destroyed), but it will play a role as to how torque is "channelled".
Have you ever noticed that 99.9% of cars/trucks/suv's that have rear wheel drive and open diff, that one (usually right) rear wheel will usually "break free" first. Factors such as leverage (axle shaft lengths) and rotational torque output come into play even assuming equal resistance with the ground.
I want to see a formula that includes the resistance being figured in.
[/ QUOTE ]
There is none for calculating torque that I know of. There is a formula that equates resistance and torque, which is along the lines of Torque must be > than the Resistance in order for work to be performed.
In a differenential, torque is neither increased or decreased due to resistance. Torque in equals torque out (assuming no parasitic losses). In a single lever design, resistance never varies the value of torque. In a system where there is 1 torque input, and 2 or more torque outputs, resistance does play a role. It does NOT decrease or increase overall torque (as energy cannot be created nor destroyed), but it will play a role as to how torque is "channelled".
Have you ever noticed that 99.9% of cars/trucks/suv's that have rear wheel drive and open diff, that one (usually right) rear wheel will usually "break free" first. Factors such as leverage (axle shaft lengths) and rotational torque output come into play even assuming equal resistance with the ground.
So resistance only comes into play when you have more than one output AND both output must be locked to turn at the same speed? I will read some more.
Seventy4Blazer
3/4 ton status
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That tells me that the rear has more power/rotation than the front.
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Or less weight. and ask any 1/4 mile guy, less weight = more power. and we are talking a few hundred pounds in most cases.
Grant
That tells me that the rear has more power/rotation than the front.
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Or less weight. and ask any 1/4 mile guy, less weight = more power. and we are talking a few hundred pounds in most cases.
Grant
Triaged
1/2 ton status
If you had a locked t-case, enough traction for the wheels to never spin, equal gearing front and rear (most of the time you can't get this...just close), and exactly the same size tire, you would have equal torque to the front and rear axle.
If your tires are all up in the air you will have the same torque to both axles.
I don't think any of those sets of conditions are of a hole lot of intrest (but the all 4 in the air sure is fun!).
If you had the front tires on some slick ice (assumed zero friction) and the rear tires on pavement, put it in reverse and steped on the gas what would happen?
All 4 tires would start to spin at the same speed, the torque on the front axle would only be the result of the inertia of the front tires/axle, the rear axle would have torque in it from the acceleration of the tires as well as the torque from the friction of the tires on the road.
If you had a 203 (not in loc) with lockers front and rear you would have the same torque going to the front and rear axles (almost none). The fronts would spin like mad and the rears would slowly move.
If your tires are all up in the air you will have the same torque to both axles.
I don't think any of those sets of conditions are of a hole lot of intrest (but the all 4 in the air sure is fun!).
If you had the front tires on some slick ice (assumed zero friction) and the rear tires on pavement, put it in reverse and steped on the gas what would happen?
All 4 tires would start to spin at the same speed, the torque on the front axle would only be the result of the inertia of the front tires/axle, the rear axle would have torque in it from the acceleration of the tires as well as the torque from the friction of the tires on the road.
If you had a 203 (not in loc) with lockers front and rear you would have the same torque going to the front and rear axles (almost none). The fronts would spin like mad and the rears would slowly move.
Triaged
1/2 ton status
Thanks. /forums/images/graemlins/thumb.gif
</font><blockquote><font class="small">In reply to:</font><hr />
If you had a 203 (not in loc) with lockers front and rear you would have the same torque going to the front and rear axles (almost none). The fronts would spin like mad and the rears would slowly move.
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"almost none"....I know mj agreed, but where does the torque go assuming the motor is twisting. Most of the torque is being applied to the axle that is slipping. The torque is "wasted" and the reason for going to fluid couplers in the newer designs. Unless torque vanishes into thin air, where does it go as energy is neither created nor destroyed with your explanation?
For example, if it takes 150 ft/lbs (at the flywheel) of torque at the rear wheels to get a 5600 lbs vehicle to accelerate at 5ft/sec squared (assuming no slip at the wheels), how do you explain the fact that in your example (and we've seen it in real life with a 203 not in loc and one set of wheels on ice) when you stomp the go pedal and the motor is making 300 ft/lbs that the vehicle will not be able to accelerate at 5 ft/sec squared? From what you say, the axle that has traction should have 1/2 the torque (150 ft/lbs), yet it cannot propel the vehicle at the rate it should.
I'm not saying you're wrong, or that I am even correct (/forums/images/graemlins/blush.gif), but this is how torque transfer has been explained to me in the past. I also know from physics classes that power can be transferred, changed, etc. but never deminishes. You could be 100% correct, but somehow it doesn't make logical sense to me on paper or in the real world experiences I have had.
After rehashing this through though, I may have to pull back the statement about the 50/50 split in a part time case, as if you have the vehicle in 4WD and only one axle has traction (say our motor had the balls to lift the front wheels), you would see close to 100% of the torque at the rear axle (if not, we would only be able to accerate 1/2 as fast when pulling the front wheels off the ground, which is not true). If ANY of this makes sense....... /forums/images/graemlins/confused.gif
All this has me rethinking my earlier posts, but even still.......maybe instead of 60/40 they should be saying ~100/~80....now even more /forums/images/graemlins/confused.gif
If you had a 203 (not in loc) with lockers front and rear you would have the same torque going to the front and rear axles (almost none). The fronts would spin like mad and the rears would slowly move.
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"almost none"....I know mj agreed, but where does the torque go assuming the motor is twisting. Most of the torque is being applied to the axle that is slipping. The torque is "wasted" and the reason for going to fluid couplers in the newer designs. Unless torque vanishes into thin air, where does it go as energy is neither created nor destroyed with your explanation?
For example, if it takes 150 ft/lbs (at the flywheel) of torque at the rear wheels to get a 5600 lbs vehicle to accelerate at 5ft/sec squared (assuming no slip at the wheels), how do you explain the fact that in your example (and we've seen it in real life with a 203 not in loc and one set of wheels on ice) when you stomp the go pedal and the motor is making 300 ft/lbs that the vehicle will not be able to accelerate at 5 ft/sec squared? From what you say, the axle that has traction should have 1/2 the torque (150 ft/lbs), yet it cannot propel the vehicle at the rate it should.
I'm not saying you're wrong, or that I am even correct (/forums/images/graemlins/blush.gif), but this is how torque transfer has been explained to me in the past. I also know from physics classes that power can be transferred, changed, etc. but never deminishes. You could be 100% correct, but somehow it doesn't make logical sense to me on paper or in the real world experiences I have had.
After rehashing this through though, I may have to pull back the statement about the 50/50 split in a part time case, as if you have the vehicle in 4WD and only one axle has traction (say our motor had the balls to lift the front wheels), you would see close to 100% of the torque at the rear axle (if not, we would only be able to accerate 1/2 as fast when pulling the front wheels off the ground, which is not true). If ANY of this makes sense....... /forums/images/graemlins/confused.gif
All this has me rethinking my earlier posts, but even still.......maybe instead of 60/40 they should be saying ~100/~80....now even more /forums/images/graemlins/confused.gif
