Calculating actual load on the torque arm?
10-18-2002, 07:19 PM
#1
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Calculating actual load on the torque arm?
A debate has come up concerning the actual load placed on the torque arm. One side lists a couple posts on LS1 Tech.com where owners have apparently broken the tailshafts off their transmissions...
http://www.ls1tech.com/ubb/cgi-bin/u...c;f=2;t=004489
http://www.ls1tech.com/ubb/cgi-bin/u...c;f=2;t=002105
He believes that at times the torque arm sees the full maximum torque (after gear multiplication) available to the engine. In other words, with a 2.66 1st gear, a 4.10 differential, and an engine with a maximum output of 400 lb-ft. of torque, he believes that the torque arm will be subjected to up to ~4,326 (400 x 2.66 x 4.1) lb-ft. of torque.
I submit that this is untrue, for several reasons. First, because there are drivetrain losses present. Second, because the torque multiplication of the ring and pinion affect the power applied to the axles only, in my opinion. Third, because even if you multiplied the available engine torque by the gear multiplication of the transmission and arrived at 1,064 lb-ft. at the output shaft (assuming no losses), you'd wad up torque arms right and left by applying that much power to the differential end of the torque arm, which would be essentially like giving it a ~3 foot breaker bar to pry on the back of the transmission with.
I think that broken output shafts are the result of hard launches on slicks, and if the clutch doesn't slip, something will eventually give way... either the tires, or if not the tires, then the U-joints or output shaft of the transmission. I believe that broken transmission tail shafts are the result of repeated fatiguing of the transmission case caused by more hard drag launches with increased power levels. It's my assertion that GM would not have mounted the torque arm to the transmission (and instead to a separate crossbrace, like the Spohn torque arm) if the torque arm saw anywhere near that kind of force.
Correct me if I'm wrong, but the torque arm's primary duties are to locate the differential correctly (for correct pinion angle) and to keep the differential from "wrapping" or twisting under power, which could cause uneven cycling of the rear suspension and wheel hop. It does see some amount of load, especially under a launch from a standing start, but I don't believe that the load placed upon it can be calculated by multiplying gear ratios by engine torque.
The reason this subject is under debate is that we're also debating the best way to mount a transmission brace and/or torque arm brace in a car which never had them, which in his case, will be required for an LS1 swap, and in mine, for an LT1 swap. In both cases, T56s are involved, and both cars are identical, having an IRS differential mounted with a bracket to a rear subframe. Stiff bushings on my car keep the differential from twisting laterally, but the "nose" of the differential can rise up and down without a Camaro-style torque arm to locate it.
My assertion is that a standard transmission brace with an integral torque arm mounting bracket bolted to the pre-existing transmission tunnel bracket mounting points (forged aluminum braces were mounted there originally) is sufficient. He asserts that welding reinforcement plates to the floor pan of the car and creating a brace which ignores the transmission tunnel brace mounting points in favor of a newly created mounting point consisting of 1/8" steel plate (like a roll bar plate, welded to the floor) and new larger bolts is necessary because of the significant load placed on the torque arm by the differential under power.
So once more, I believe the torque arm sees some load, especially on launch, but not a huge load and nowhere near "maximum torque" after gear multiplication. He believes that the torque arm sees everything the engine can dish out. If anything, I'd guess that the answer is somewhere in the middle, but how would one calculate the actual strain on the torque arm? Any help in resolving this either way would be greatly appreciated.
Thanks!
http://www.ls1tech.com/ubb/cgi-bin/u...c;f=2;t=004489
http://www.ls1tech.com/ubb/cgi-bin/u...c;f=2;t=002105
He believes that at times the torque arm sees the full maximum torque (after gear multiplication) available to the engine. In other words, with a 2.66 1st gear, a 4.10 differential, and an engine with a maximum output of 400 lb-ft. of torque, he believes that the torque arm will be subjected to up to ~4,326 (400 x 2.66 x 4.1) lb-ft. of torque.
I submit that this is untrue, for several reasons. First, because there are drivetrain losses present. Second, because the torque multiplication of the ring and pinion affect the power applied to the axles only, in my opinion. Third, because even if you multiplied the available engine torque by the gear multiplication of the transmission and arrived at 1,064 lb-ft. at the output shaft (assuming no losses), you'd wad up torque arms right and left by applying that much power to the differential end of the torque arm, which would be essentially like giving it a ~3 foot breaker bar to pry on the back of the transmission with.
I think that broken output shafts are the result of hard launches on slicks, and if the clutch doesn't slip, something will eventually give way... either the tires, or if not the tires, then the U-joints or output shaft of the transmission. I believe that broken transmission tail shafts are the result of repeated fatiguing of the transmission case caused by more hard drag launches with increased power levels. It's my assertion that GM would not have mounted the torque arm to the transmission (and instead to a separate crossbrace, like the Spohn torque arm) if the torque arm saw anywhere near that kind of force.
Correct me if I'm wrong, but the torque arm's primary duties are to locate the differential correctly (for correct pinion angle) and to keep the differential from "wrapping" or twisting under power, which could cause uneven cycling of the rear suspension and wheel hop. It does see some amount of load, especially under a launch from a standing start, but I don't believe that the load placed upon it can be calculated by multiplying gear ratios by engine torque.
The reason this subject is under debate is that we're also debating the best way to mount a transmission brace and/or torque arm brace in a car which never had them, which in his case, will be required for an LS1 swap, and in mine, for an LT1 swap. In both cases, T56s are involved, and both cars are identical, having an IRS differential mounted with a bracket to a rear subframe. Stiff bushings on my car keep the differential from twisting laterally, but the "nose" of the differential can rise up and down without a Camaro-style torque arm to locate it.
My assertion is that a standard transmission brace with an integral torque arm mounting bracket bolted to the pre-existing transmission tunnel bracket mounting points (forged aluminum braces were mounted there originally) is sufficient. He asserts that welding reinforcement plates to the floor pan of the car and creating a brace which ignores the transmission tunnel brace mounting points in favor of a newly created mounting point consisting of 1/8" steel plate (like a roll bar plate, welded to the floor) and new larger bolts is necessary because of the significant load placed on the torque arm by the differential under power.
So once more, I believe the torque arm sees some load, especially on launch, but not a huge load and nowhere near "maximum torque" after gear multiplication. He believes that the torque arm sees everything the engine can dish out. If anything, I'd guess that the answer is somewhere in the middle, but how would one calculate the actual strain on the torque arm? Any help in resolving this either way would be greatly appreciated.
Thanks!
10-19-2002, 09:57 AM
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Re: Calculating actual load on the torque arm?
Originally posted by jimlab
A debate has come up concerning the actual load placed on the torque arm. One side... believes that at times the torque arm sees the full maximum torque (after gear multiplication) available to the engine. In other words, with a 2.66 1st gear, a 4.10 differential, and an engine with a maximum output of 400 lb-ft. of torque, he believes that the torque arm will be subjected to up to ~4,326 (400 x 2.66 x 4.1) lb-ft. of torque.
I submit that this is untrue, for several reasons. First, because there are drivetrain losses present. Second, because the torque multiplication of the ring and pinion affect the power applied to the axles only, in my opinion. Third, because even if you multiplied the available engine torque by the gear multiplication of the transmission and arrived at 1,064 lb-ft. at the output shaft (assuming no losses), you'd wad up torque arms right and left by applying that much power to the differential end of the torque arm, which would be essentially like giving it a ~3 foot breaker bar to pry on the back of the transmission with.
I think that broken output shafts are the result of hard launches on slicks, and if the clutch doesn't slip, something will eventually give way... either the tires, or if not the tires, then the U-joints or output shaft of the transmission. I believe that broken transmission tail shafts are the result of repeated fatiguing of the transmission case caused by more hard drag launches with increased power levels. It's my assertion that GM would not have mounted the torque arm to the transmission (and instead to a separate crossbrace, like the Spohn torque arm) if the torque arm saw anywhere near that kind of force.
Correct me if I'm wrong, but the torque arm's primary duties are to locate the differential correctly (for correct pinion angle) and to keep the differential from "wrapping" or twisting under power, which could cause uneven cycling of the rear suspension and wheel hop. It does see some amount of load, especially under a launch from a standing start, but I don't believe that the load placed upon it can be calculated by multiplying gear ratios by engine torque.
The reason this subject is under debate is that we're also debating the best way to mount a transmission brace and/or torque arm brace in a car which never had them, which in his case, will be required for an LS1 swap, and in mine, for an LT1 swap. In both cases, T56s are involved, and both cars are identical, having an IRS differential mounted with a bracket to a rear subframe. Stiff bushings on my car keep the differential from twisting laterally, but the "nose" of the differential can rise up and down without a Camaro-style torque arm to locate it.
My assertion is that a standard transmission brace with an integral torque arm mounting bracket bolted to the pre-existing transmission tunnel bracket mounting points (forged aluminum braces were mounted there originally) is sufficient. He asserts that welding reinforcement plates to the floor pan of the car and creating a brace which ignores the transmission tunnel brace mounting points in favor of a newly created mounting point consisting of 1/8" steel plate (like a roll bar plate, welded to the floor) and new larger bolts is necessary because of the significant load placed on the torque arm by the differential under power.
So once more, I believe the torque arm sees some load, especially on launch, but not a huge load and nowhere near "maximum torque" after gear multiplication. He believes that the torque arm sees everything the engine can dish out. If anything, I'd guess that the answer is somewhere in the middle, but how would one calculate the actual strain on the torque arm? Any help in resolving this either way would be greatly appreciated.
Thanks!
A debate has come up concerning the actual load placed on the torque arm. One side... believes that at times the torque arm sees the full maximum torque (after gear multiplication) available to the engine. In other words, with a 2.66 1st gear, a 4.10 differential, and an engine with a maximum output of 400 lb-ft. of torque, he believes that the torque arm will be subjected to up to ~4,326 (400 x 2.66 x 4.1) lb-ft. of torque.
I submit that this is untrue, for several reasons. First, because there are drivetrain losses present. Second, because the torque multiplication of the ring and pinion affect the power applied to the axles only, in my opinion. Third, because even if you multiplied the available engine torque by the gear multiplication of the transmission and arrived at 1,064 lb-ft. at the output shaft (assuming no losses), you'd wad up torque arms right and left by applying that much power to the differential end of the torque arm, which would be essentially like giving it a ~3 foot breaker bar to pry on the back of the transmission with.
I think that broken output shafts are the result of hard launches on slicks, and if the clutch doesn't slip, something will eventually give way... either the tires, or if not the tires, then the U-joints or output shaft of the transmission. I believe that broken transmission tail shafts are the result of repeated fatiguing of the transmission case caused by more hard drag launches with increased power levels. It's my assertion that GM would not have mounted the torque arm to the transmission (and instead to a separate crossbrace, like the Spohn torque arm) if the torque arm saw anywhere near that kind of force.
Correct me if I'm wrong, but the torque arm's primary duties are to locate the differential correctly (for correct pinion angle) and to keep the differential from "wrapping" or twisting under power, which could cause uneven cycling of the rear suspension and wheel hop. It does see some amount of load, especially under a launch from a standing start, but I don't believe that the load placed upon it can be calculated by multiplying gear ratios by engine torque.
The reason this subject is under debate is that we're also debating the best way to mount a transmission brace and/or torque arm brace in a car which never had them, which in his case, will be required for an LS1 swap, and in mine, for an LT1 swap. In both cases, T56s are involved, and both cars are identical, having an IRS differential mounted with a bracket to a rear subframe. Stiff bushings on my car keep the differential from twisting laterally, but the "nose" of the differential can rise up and down without a Camaro-style torque arm to locate it.
My assertion is that a standard transmission brace with an integral torque arm mounting bracket bolted to the pre-existing transmission tunnel bracket mounting points (forged aluminum braces were mounted there originally) is sufficient. He asserts that welding reinforcement plates to the floor pan of the car and creating a brace which ignores the transmission tunnel brace mounting points in favor of a newly created mounting point consisting of 1/8" steel plate (like a roll bar plate, welded to the floor) and new larger bolts is necessary because of the significant load placed on the torque arm by the differential under power.
So once more, I believe the torque arm sees some load, especially on launch, but not a huge load and nowhere near "maximum torque" after gear multiplication. He believes that the torque arm sees everything the engine can dish out. If anything, I'd guess that the answer is somewhere in the middle, but how would one calculate the actual strain on the torque arm? Any help in resolving this either way would be greatly appreciated.
Thanks!
Yes, the torque arm feels ALL of the torque applied to the rear axle, but it may be worse than you think. With a high-stall torque converter, engine torque is multiplied by an addition factor of 2 (or more) at launch.
Of course the limiting factor is tire traction; only that torque that actually gets to the ground needs to be reacted. For an example, let's say a Camaro with super sticky slicks, 400 lb-ft @ 4000 rpm and a 4000 rpm stall convertor in an A4 and 4.10's. What's the torque applied to the axles if the tires don't slip?
400 x 2.2 (converter) x 3.06 (A4 first gear) x 4.10 = 11,040 ft-lb!
Sure drivetrain losses are present, so let's say the net is 10,000 ft-lb.
Even with an M6, if you have the traction available, and sidestep (or 'unfoot' as some Brits say) the clutch at 4000+, the momentum stored in the rotating engine bits and fllywheel is released into the system quickly as a torque spike (like you see on a chassis dyno at a shift point) and the loads can approach those of the A4 with a high stall TC.
You are correct that the torque arm's job is to keep the axle housing from wrapping up. It's the only thing preventing that from happening in the F-body rear suspension.
Now, assuming the torque arm is about 5 feet long, and rounding the maximum axle torque to 10,000 ft-lb, the front mounting point of the torque arm sees a vertical load of 10,000/5 = 2000 lbs. Now that a ton of load (pun intended) for the stock mounting, which would maybe see less than 1/2 of that with stock tires/engine/converter. Remember, the stock tire traction is the weak point or safety link in the chain. It's the slicks' ability to put the power down that loads the suspension so much more than the OEM design.
I agree that broken torque arm mounts are from repeated high loading (fatigue failure). Usually fatigue failures are from thousands or tens of thousands of cycles, but with enough load, a few cycles will do it. If the ultimate strength of a part is exceeded the very first time it is loaded it will fail on it's first loading.
I have to side with your adversary: the torque arm absolutely sees it all. For every action there is an equal an opposite reaction...that's the torque arm's job. The answer is NOT "somewhere in the middle", unfortunately.
Wherever the torque arm mounting point, it must be strong enough to take that load. BTW, shortening the arm to 3 feet bumps the load up to over 3300 lbs. in my example. Like almost anything else in engineering of a machine, when you have higher loads, you need stronger parts, and when you beef up the weakest, the failure progresses to the next weaker point. For example: rear end gears, axle shafts, torque arm mount, drive shaft, u-joints, trans gears, clutch but not necessarily in that order.
My $.02.
10-19-2002, 10:43 AM
#3
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Cant help you with the actual loads placed on the arms but if I read right you are using independant rear suspention? I'm guessing vette? I wouldnt think you would need to go with a torque arm setup but some thing a little more solid since you dont have to worry about travel.
http://www.progressiveautomotive.com/sweetrear.htm
http://www.progressiveautomotive.com/sweetrear.htm
10-19-2002, 05:03 PM
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Re: Re: Calculating actual load on the torque arm?
Originally posted by OldSStroker
Now, assuming the torque arm is about 5 feet long, and rounding the maximum axle torque to 10,000 ft-lb, the front mounting point of the torque arm sees a vertical load of 10,000/5 = 2000 lbs. Now that a ton of load (pun intended) for the stock mounting, which would maybe see less than 1/2 of that with stock tires/engine/converter. Remember, the stock tire traction is the weak point or safety link in the chain. It's the slicks' ability to put the power down that loads the suspension so much more than the OEM design.
< Snip >
I have to side with your adversary: the torque arm absolutely sees it all. For every action there is an equal an opposite reaction...that's the torque arm's job. The answer is NOT "somewhere in the middle", unfortunately.
Now, assuming the torque arm is about 5 feet long, and rounding the maximum axle torque to 10,000 ft-lb, the front mounting point of the torque arm sees a vertical load of 10,000/5 = 2000 lbs. Now that a ton of load (pun intended) for the stock mounting, which would maybe see less than 1/2 of that with stock tires/engine/converter. Remember, the stock tire traction is the weak point or safety link in the chain. It's the slicks' ability to put the power down that loads the suspension so much more than the OEM design.
< Snip >
I have to side with your adversary: the torque arm absolutely sees it all. For every action there is an equal an opposite reaction...that's the torque arm's job. The answer is NOT "somewhere in the middle", unfortunately.
Wherever the torque arm mounting point, it must be strong enough to take that load. BTW, shortening the arm to 3 feet bumps the load up to over 3300 lbs. in my example.
Like almost anything else in engineering of a machine, when you have higher loads, you need stronger parts, and when you beef up the weakest, the failure progresses to the next weaker point. For example: rear end gears, axle shafts, torque arm mount, drive shaft, u-joints, trans gears, clutch but not necessarily in that order.
My only reluctance in welding and drilling the floor pan is that my car only has 13,000 miles on it and is in nearly new condition. The goal was also to produce a truly bolt-on solution for others who want to swap engines and transmissions that didn't require drilling, cutting or welding. My initial plan was to use a boxed "frame" using 4 existing mounting points instead of only 2, and using that for the basis of the trans mount and torque arm mount. I'll have to take another look at the area and determine whether or not that's feasible. It might even be possible to use 6 mounting points, since there were originally 3 forged aluminum braces on the transmission tunnel.
96ltz, the car is a 1995 Mazda RX-7. It already has an excellent IRS setup. The weak points are the axles (easily remedied) and the tripod CV cups and stub axles (not so easily remedied) and the requirement for some sort of torque arm to replace the Power Plant Frame (PPF) which originally linked the transmission to the differential. There was no transmission mount originally, it being suspended between the engine and differential supported by a rigid boxed "C" shaped (in profile) brace. The differential is supported at the rear of the car by its bracket and two large bushings, and the engine was supported by its mounts in front. An F-body style adjustable torque arm seemed like just the ticket to support and align the differential properly in the absence of the PPF.
10-19-2002, 05:30 PM
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Jim,
C4 Corvettes used the same type of link between the transmission and the rearend, but they happen to use a ZF trans. Unless you are stuck on using the T56, it may provide another avenue to use the ZF, and fabricate a new "C" channel to mate the ZF with the Mazda IRS.
Another option would be to use a type of adjustable pinion snubber like the old Mopars used. It just places a preload on the front part of the third member, just behind the yoke. This would eliiminate any, or all "wrapping" of the IRS, and would be much more simple than trying to fabricate a new torque arm, or even the "C" channel method I mentioned above.
I personally believe the only reason a torque arm is needed, is when the differential actually has to move up and down with suspension movements. With an IRS setup, the diff. doesn't need to move at all, so why are you trying to engineer a pivot for it instead of just making it solid like it was originally designed for. That is, unless I'm totally missing the point completely. JMHO's.
Mike
BTW, the block I bought from ya is going to the machine shop in about 2 weeks.
C4 Corvettes used the same type of link between the transmission and the rearend, but they happen to use a ZF trans. Unless you are stuck on using the T56, it may provide another avenue to use the ZF, and fabricate a new "C" channel to mate the ZF with the Mazda IRS.
Another option would be to use a type of adjustable pinion snubber like the old Mopars used. It just places a preload on the front part of the third member, just behind the yoke. This would eliiminate any, or all "wrapping" of the IRS, and would be much more simple than trying to fabricate a new torque arm, or even the "C" channel method I mentioned above.
I personally believe the only reason a torque arm is needed, is when the differential actually has to move up and down with suspension movements. With an IRS setup, the diff. doesn't need to move at all, so why are you trying to engineer a pivot for it instead of just making it solid like it was originally designed for. That is, unless I'm totally missing the point completely. JMHO's.
Mike
BTW, the block I bought from ya is going to the machine shop in about 2 weeks.
10-19-2002, 05:32 PM
#6
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Location: TN
Posts: 243
If I'm picturing this right the rear is joined sort of like a trans/transfer case? What kind of coupling did it have? How did you convert it to a yolk if thats what you are using?
Could you use some of the existing holes to fab up a crossmember like in the link above?
Could you use some of the existing holes to fab up a crossmember like in the link above?
10-19-2002, 11:19 PM
#7
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Posts: 4
math
The only true way to find out the trq arm load is to connect the front to a load cell(trq registering block) and moniter it during a during a full pass down the track. And as for the trq numbers aplied to the trans it`s`alot lower than you think. because there are a some things you are not thinking about. first is rolling resistace. second is the cars weight. last is the length of the trq arm. when you think about trq arm think of it like a breaker bar, the socket end being the rear end and the handle being the front mount. the longer the bar the less trq will be aplied to the mount.
Chris S
94 Z28 A4 13.96@98
72 MAVRICK GRABBER 12.53@104 (1.67-60`)
Chris S
94 Z28 A4 13.96@98
72 MAVRICK GRABBER 12.53@104 (1.67-60`)
10-20-2002, 02:20 AM
#8
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Originally posted by Mikey 97Z M6
C4 Corvettes used the same type of link between the transmission and the rearend, but they happen to use a ZF trans. Unless you are stuck on using the T56, it may provide another avenue to use the ZF, and fabricate a new "C" channel to mate the ZF with the Mazda IRS.
C4 Corvettes used the same type of link between the transmission and the rearend, but they happen to use a ZF trans. Unless you are stuck on using the T56, it may provide another avenue to use the ZF, and fabricate a new "C" channel to mate the ZF with the Mazda IRS.
Another option would be to use a type of adjustable pinion snubber like the old Mopars used. It just places a preload on the front part of the third member, just behind the yoke. This would eliiminate any, or all "wrapping" of the IRS, and would be much more simple than trying to fabricate a new torque arm, or even the "C" channel method I mentioned above.
The 3rd gen. RX-7 rear suspension looks like this...
http://home.earthlink.net/~jimlab/pi...suspension.jpg
The black subframe bolts to the body of the car, and the rear suspension components are mounted on it. The four nuts on the top rear of the differential housing (slightly orange) are for attaching the differential mount, which (although difficult to see in this picture) is already mounted on the subframe and is located immediately behind the differential. On either end of the mount, you may be able to pick out the bushings. These are Nylon 6-6 replacements and are dyed a hot pink. You can see just the outside edge of the bushings in the picture above, and again, I apologize for the lack of clarity.
The differential will swing freely up and down when attached to this mount, and it was the power plant frame (PPF) which bolted to the front of the housing (you can see the top of the upper two studs... with a slight green tint on the front of the differential housing, on the right hand side) using four 17mm studs. This is where the torque arm (that has already been developed) would fasten to the differential housing, and the front point of the arm would attach to the transmission brace.
In the next picture, you can see the type of transmission tunnel bracing I'm talking about...
http://home.earthlink.net/~jimlab/pi...peed_brace.jpg
This is the Mazdaspeed brace and you can ignore the front "V" shaped brace, since it doesn't apply to my swap. The rear brace, which is a long rectangle with two extensions bolts to the three OEM transmission tunnel brace mounting points that I was talking about in my last post. These braces were intended to make the chassis more rigid (as were an integrated rear strut tower brace and an optional front strut tower brace). The OEM braces are forged aluminum, but this aftermarket solution is more than likely steel. Tying the six mounting points together is probably an improvement, and it was this type of configuration, with a bracket for the transmission mount, a forward mounting point for a torque arm, and an integrated driveshaft loop that I was thinking of. The other guy is considering tying into the floor pan and welding in reinforcement plates for new mounting points.
If there were no need for a torque arm, then the stock bracing could be used in the rearmost two positions, and a new transmission mount/brace could be fabricated to bolt to the front position, since the trans mount doesn't see that much abuse, unless I'm mistaken.
I personally believe the only reason a torque arm is needed, is when the differential actually has to move up and down with suspension movements. With an IRS setup, the diff. doesn't need to move at all, so why are you trying to engineer a pivot for it instead of just making it solid like it was originally designed for.
Anyway, something could be fabricated to locate the differential pinion angle properly, since the differential really doesn't need to move up or down if it has been positioned properly. I'm open to any suggestions.
BTW, the block I bought from ya is going to the machine shop in about 2 weeks.
10-20-2002, 03:00 AM
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Here are a couple more pictures that might illustrate what I'm working with a little better...
Here's a clearer (but smaller) picture of the rear suspension of the 3rd gen. (1993-1995) RX-7...
http://beyondtheredline.fd3s.net/images/93_rearsusp.jpg
The next picture shows one of the OEM forged aluminum trans tunnel braces that I've been talking about, and at the very top center of the picture, you can see the front of the black power plant frame (PPF) where it attaches to the rear of the Mazda 5-speed.
http://www.negative-camber.org/crispyrx7/begin.jpg
The next picture shows the differential housing from below, with the PPF bolted to it. The rear most aluminum trans tunnel brace actually passes through the rear of the PPF. As you can see, the driveline is tucked into the inner portion of the "C" shape of the PPF.
http://www.negative-camber.org/crispyrx7/begindiff.jpg
Here's a shot from the front showing the differential housing with the driveline removed, but the PPF still attached. The green piece is one of the tripod "cups" which forms the inner CV joint of the rear axle system. It has a "stub" axle which inserts into (and locks) the differential itself, which is a Torsen limited slip unit in the stock configuration.
Here's a great side view of the PPF, hanging at the front where it would bolt to the transmission. You can see why I thought that a torque arm was a logical replacement, since the configuration is similar, although a torque arm obviously does not support the rear of the transmission.
http://www.negative-camber.org/crispyrx7/ppfangle.jpg
And here's a side view of the differential housing, showing the PPF and the aluminum brace (going through the PPF). You'll have to drag and drop this one to your browser's address box, or to another browser window.
http://cp_www.tripod.com/rotary/images/pg32_18b.jpg
There's a picture of the entire drivetrain out of the car, from engine to differential, with the transmission and PPF suspended in between, and complete front and rear suspensions floating around somewhere, but unfortunately I wasn't able to find it. Hopefully the pictures above clear up the position of the parts that I'm talking about and the configuration of the differential and rear suspension in the 3rd gen. RX-7.
Here's a clearer (but smaller) picture of the rear suspension of the 3rd gen. (1993-1995) RX-7...
http://beyondtheredline.fd3s.net/images/93_rearsusp.jpg
The next picture shows one of the OEM forged aluminum trans tunnel braces that I've been talking about, and at the very top center of the picture, you can see the front of the black power plant frame (PPF) where it attaches to the rear of the Mazda 5-speed.
http://www.negative-camber.org/crispyrx7/begin.jpg
The next picture shows the differential housing from below, with the PPF bolted to it. The rear most aluminum trans tunnel brace actually passes through the rear of the PPF. As you can see, the driveline is tucked into the inner portion of the "C" shape of the PPF.
http://www.negative-camber.org/crispyrx7/begindiff.jpg
Here's a shot from the front showing the differential housing with the driveline removed, but the PPF still attached. The green piece is one of the tripod "cups" which forms the inner CV joint of the rear axle system. It has a "stub" axle which inserts into (and locks) the differential itself, which is a Torsen limited slip unit in the stock configuration.
Here's a great side view of the PPF, hanging at the front where it would bolt to the transmission. You can see why I thought that a torque arm was a logical replacement, since the configuration is similar, although a torque arm obviously does not support the rear of the transmission.
http://www.negative-camber.org/crispyrx7/ppfangle.jpg
And here's a side view of the differential housing, showing the PPF and the aluminum brace (going through the PPF). You'll have to drag and drop this one to your browser's address box, or to another browser window.
http://cp_www.tripod.com/rotary/images/pg32_18b.jpg
There's a picture of the entire drivetrain out of the car, from engine to differential, with the transmission and PPF suspended in between, and complete front and rear suspensions floating around somewhere, but unfortunately I wasn't able to find it. Hopefully the pictures above clear up the position of the parts that I'm talking about and the configuration of the differential and rear suspension in the 3rd gen. RX-7.
10-20-2002, 09:02 AM
#10
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Re: Re: Re: Calculating actual load on the torque arm?
Originally posted by jimlab
Yet you just stated that you believe the load on the nose of the torque arm to be around 1/5th of the "maximum torque" available with gear multiplication... which is quite a big difference. My contention was that the nose did not see maximum engine torque, and you seem to be agreeing with me, at least on that point.
I used three feet as a rough estimate. The car is shorter than an F-body, I believe, and didn't originally have a T56 mounted in it, so I'm not sure what length the torque arm ends up being. I wasn't aware that the F-body torque arms were so long, never having seen one in person.
Yet you just stated that you believe the load on the nose of the torque arm to be around 1/5th of the "maximum torque" available with gear multiplication... which is quite a big difference. My contention was that the nose did not see maximum engine torque, and you seem to be agreeing with me, at least on that point.
I used three feet as a rough estimate. The car is shorter than an F-body, I believe, and didn't originally have a T56 mounted in it, so I'm not sure what length the torque arm ends up being. I wasn't aware that the F-body torque arms were so long, never having seen one in person.
Torque is load (or FORCE) times distance. As others mentioned, think of a breaker bar or torque wrench. The front torque arm mount is your hand resisting the torque applied at the socket (axle). The longer the lever arm, the less the force. 10,000 ft-lb at 5 ft = 2000 lb. 10,000 ft-lb at 3 ft = 3333 lb., and 10,000 ft-lb with a 100 ft. long torque arm would only be 100 lb. Your front mount will need to resist that kind of load, which depends on the torque arm length.
The C4 Vette ITS used a similar torque arm to yours.
FWIW, here's a long, boring description:
A vehicle is accelerated by a FORCE pushing it forward. Where does this force come from?
The torque applied to the drive axles tries to twist the wheels and tires. If the tire has a 24 inch (2 ft) diameter, or a 1 ft. radius, the FORCE at the road (with no slippage) is the axle torque (ft-lb) divided by the tire radius (ft), or in my original example, 10,000 ft-lb / 1 ft = 10,000 lb. of FORCE at the road/tire interface trying to push the road (connected to the Earth) back away from the car. Since the Earth is considerably more massive than even my Impala SS, it moves very little, but the car moves forward a lot more.
Newton is credited with saying "For every action there is an equal and opposite reaction". That's how things work.
That FORCE at the road has a equal reaction FORCE, but opposite in direction, applied to the car parallel to the road where the wheel is supported (the outer end of the solid axle or the outer bearing carrier in a IRS). This FORCE is then transferred to the chassis through suspension links. In a F-body, it's the two lower control arms. In a C4 Vette it's the 4 trailing arms. In a double wishbone (A-arm) IRS it's through the A-arms. That's what accelerates the vehicle.
But, there's more. All that torque applied to the drive axles also has to be resisted. It's similar to using your electric drill to bore a hole in piece of wood. If you hit a hard spot the drill motor tends to twist your hand opposite the direction of the drill bit. your hand/arm supply the torque reaction. Similarly, imagine the pinion gear trying to turn the ring gear which is basically stationary because the slicks bite so well. The pinion actually tries to climb up the ring gear. Because it is restrained in the housing, the housing rotates. That's the torque reaction which must be resisted by the suspension.
Bottom line is, except for driveline losses due to friction, etc. ALL of the torque applied to the drive axles has to be resisted by the suspension bits. With a simple torque arm, the FORCE or load at the forward mounting point varies directly with the length of the arm.
More than you probably wanted to hear.
10-20-2002, 01:54 PM
#11
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Re: Re: Re: Re: Calculating actual load on the torque arm?
Originally posted by OldSStroker
Torque is load (or FORCE) times distance. As others mentioned, think of a breaker bar or torque wrench. The front torque arm mount is your hand resisting the torque applied at the socket (axle). The longer the lever arm, the less the force. 10,000 ft-lb at 5 ft = 2000 lb. 10,000 ft-lb at 3 ft = 3333 lb., and 10,000 ft-lb with a 100 ft. long torque arm would only be 100 lb. Your front mount will need to resist that kind of load, which depends on the torque arm length.
Torque is load (or FORCE) times distance. As others mentioned, think of a breaker bar or torque wrench. The front torque arm mount is your hand resisting the torque applied at the socket (axle). The longer the lever arm, the less the force. 10,000 ft-lb at 5 ft = 2000 lb. 10,000 ft-lb at 3 ft = 3333 lb., and 10,000 ft-lb with a 100 ft. long torque arm would only be 100 lb. Your front mount will need to resist that kind of load, which depends on the torque arm length.
A 1 lb. weight hanging from the end of a 1-foot long bar is exerting 1 lb-ft. of torque. A 2lb. weight would be 2 lb-ft. and so on. Hanging the same 1 lb. weight from the end of a 2-foot long bar is equivalent to 2 lb-ft. On the end of a 3-foot bar, 3 lb-ft. That's why a breaker bar will often allow you to loosen a very tight bolt where you couldn't hope to budge it with a standard length ratchet handle, let alone an open-end wrench which might only have an overall length of 5-7 inches.
So how is exerting torque or force on the end of a 5-foot (for sake of arguement) torque arm connected to a single forward mounting point reducing the torque seen at the mounting point? If the mounting point were fixed, I'd expect the action on the torque arm to be equivalent to giving the differential a 5-foot breaker bar and letting it twist on the mount. But the front mount is not fixed, nor is the angle of rotation through the normal movement range of the differential very large, so I would expect that torque measured at the nose of the torque arm would be quite a bit higher than flywheel torque, but certainly less than the maximum torque available through gear multiplication.
I suppose that the only real way to solve this would be to, as 1320 MAN suggested, attach a device to the front of the torque arm to actually measure the strain under which it is placed. Or contact someone like Spohn, who may have practical knowledge (having developed a replacement part based on some presumably observed and quantified shortcoming of the original mounting system) of the level of strain applied to the nose of the torque arm.
Or just forget about using a torque arm altogether, since the link provided by 96ltz shows Corvette IRS configurations with the front of the differential fix-mounted (presumably with a bushing for noise isolation) to a bar either above or below the differential. It doesn't appear that except for properly setting pinion angle, that it is very critical that the differential in an IRS configuration, at least one such as the C4's or mine, be able to rotate during normal operation.
10-20-2002, 02:30 PM
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Re: Re: Re: Re: Re: Calculating actual load on the torque arm?
Originally posted by jimlab
True, but that's the second time someone has said that the length of the torque arm reduces torque, which is completely opposite of what I'd expect.
A 1 lb. weight hanging from the end of a 1-foot long bar is exerting 1 lb-ft. of torque. A 2lb. weight would be 2 lb-ft. and so on. Hanging the same 1 lb. weight from the end of a 2-foot long bar is equivalent to 2 lb-ft. On the end of a 3-foot bar, 3 lb-ft. That's why a breaker bar will often allow you to loosen a very tight bolt where you couldn't hope to budge it with a standard length ratchet handle, let alone an open-end wrench which might only have an overall length of 5-7 inches.
So how is exerting torque or force on the end of a 5-foot (for sake of arguement) torque arm connected to a single forward mounting point reducing the torque seen at the mounting point? If the mounting point were fixed, I'd expect the action on the torque arm to be equivalent to giving the differential a 5-foot breaker bar and letting it twist on the mount. But the front mount is not fixed, nor is the angle of rotation through the normal movement range of the differential very large, so I would expect that torque measured at the nose of the torque arm would be quite a bit higher than flywheel torque, but certainly less than the maximum torque available through gear multiplication.
Or just forget about using a torque arm altogether, since the link provided by 96ltz shows Corvette IRS configurations with the front of the differential fix-mounted (presumably with a bushing for noise isolation) to a bar either above or below the differential. It doesn't appear that except for properly setting pinion angle, that it is very critical that the differential in an IRS configuration, at least one such as the C4's or mine, be able to rotate during normal operation.
True, but that's the second time someone has said that the length of the torque arm reduces torque, which is completely opposite of what I'd expect.
A 1 lb. weight hanging from the end of a 1-foot long bar is exerting 1 lb-ft. of torque. A 2lb. weight would be 2 lb-ft. and so on. Hanging the same 1 lb. weight from the end of a 2-foot long bar is equivalent to 2 lb-ft. On the end of a 3-foot bar, 3 lb-ft. That's why a breaker bar will often allow you to loosen a very tight bolt where you couldn't hope to budge it with a standard length ratchet handle, let alone an open-end wrench which might only have an overall length of 5-7 inches.
So how is exerting torque or force on the end of a 5-foot (for sake of arguement) torque arm connected to a single forward mounting point reducing the torque seen at the mounting point? If the mounting point were fixed, I'd expect the action on the torque arm to be equivalent to giving the differential a 5-foot breaker bar and letting it twist on the mount. But the front mount is not fixed, nor is the angle of rotation through the normal movement range of the differential very large, so I would expect that torque measured at the nose of the torque arm would be quite a bit higher than flywheel torque, but certainly less than the maximum torque available through gear multiplication.
Or just forget about using a torque arm altogether, since the link provided by 96ltz shows Corvette IRS configurations with the front of the differential fix-mounted (presumably with a bushing for noise isolation) to a bar either above or below the differential. It doesn't appear that except for properly setting pinion angle, that it is very critical that the differential in an IRS configuration, at least one such as the C4's or mine, be able to rotate during normal operation.
Jim,
Not to flame you, but I think you are missing the difference between torque (force x distance) and force (load) alone when it comes to the torque arm loads. The longer torque arm (TA) doesn't change TORQUE, it changes the FORCE or LOAD at the front end of the arm. There is NO torque at the front end of the TA, just a vertical LOAD acting over a distance (the TA length) which is the torque felt at the axle.
Your breaker bar explanation is perfect. Just apply that to the torque arm. Your hand on the handle of the breaker bar is the front torque arm mount. As the bar gets longer, you have to push with less force to resist a given torque. The forward mounting point of the TA doesn't feel a torque (twisting), it only feels a vertical load, just like your hand on the bar.
Effectively, the front TA mount IS fixed, and the only movement due to resisting the torque is bushing deflection and slight bending of the TA. Neither changes the loads applied to the front mount more than a whisker. If you buy the torque reaction explanation that tries to rotate the axle housing in the opposite direction the wheels are turning, then the TA is just the breaker bar resisting this torque.
Torque arm length and mounting position gets even more complicated because it affects the 'anti squat' of the rear suspension. Anti squat allows the torque reaction of the rear suspension to actually push down on the rear tires for more traction. If you've ever seen a drag car's rear end LIFT during launch, rather than settle (squat), you can see the result of anti squat. How that works is way more complex than just the torque arm loads, so this probably isn't the thread to discuss it.
That bar on the Vette ('84-'96) differential IS a torque arm, and, no, it doesn't rotate. The only reason the f-car (solid axle) TA moves is for suspension travel up and down.
I may just be confusing you. That's not my intent.
Good luck.
10-20-2002, 02:48 PM
#13
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On an fbod, force is transfered via the TA and the LCA. If the LCA were one the same plane as the the rear axle, the the TA would bear the grunt of it all. Still not as bad as you think since the TA length is so long, I still like not blowing out the Tranny mount with my traclink since it has an alternate mounting point.
10-20-2002, 03:11 PM
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Re: Re: Re: Re: Re: Re: Calculating actual load on the torque arm?
Originally posted by OldSStroker
Jim,
Not to flame you, but I think you are missing the difference between torque (force x distance) and force (load) alone when it comes to the torque arm loads. The longer torque arm (TA) doesn't change TORQUE, it changes the FORCE or LOAD at the front end of the arm. There is NO torque at the front end of the TA, just a vertical LOAD acting over a distance (the TA length) which is the torque felt at the axle.
Your breaker bar explanation is perfect. Just apply that to the torque arm. Your hand on the handle of the breaker bar is the front torque arm mount. As the bar gets longer, you have to push with less force to resist a given torque. The forward mounting point of the TA doesn't feel a torque (twisting), it only feels a vertical load, just like your hand on the bar.
Jim,
Not to flame you, but I think you are missing the difference between torque (force x distance) and force (load) alone when it comes to the torque arm loads. The longer torque arm (TA) doesn't change TORQUE, it changes the FORCE or LOAD at the front end of the arm. There is NO torque at the front end of the TA, just a vertical LOAD acting over a distance (the TA length) which is the torque felt at the axle.
Your breaker bar explanation is perfect. Just apply that to the torque arm. Your hand on the handle of the breaker bar is the front torque arm mount. As the bar gets longer, you have to push with less force to resist a given torque. The forward mounting point of the TA doesn't feel a torque (twisting), it only feels a vertical load, just like your hand on the bar.
That bar on the Vette ('84-'96) differential IS a torque arm, and, no, it doesn't rotate. The only reason the f-car (solid axle) TA moves is for suspension travel up and down.
Thanks!
10-20-2002, 04:02 PM
#15
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Re: Re: Re: Re: Re: Calculating actual load on the torque arm?
Originally posted by jimlab
Or just forget about using a torque arm altogether, since the link provided by 96ltz shows Corvette IRS configurations with the front of the differential fix-mounted (presumably with a bushing for noise isolation) to a bar either above or below the differential. It doesn't appear that except for properly setting pinion angle, that it is very critical that the differential in an IRS configuration, at least one such as the C4's or mine, be able to rotate during normal operation.
Or just forget about using a torque arm altogether, since the link provided by 96ltz shows Corvette IRS configurations with the front of the differential fix-mounted (presumably with a bushing for noise isolation) to a bar either above or below the differential. It doesn't appear that except for properly setting pinion angle, that it is very critical that the differential in an IRS configuration, at least one such as the C4's or mine, be able to rotate during normal operation.
It looks like from the first picture you put up you triangulate off of the front crossmember (the one that doesnt meet) as long as nothing is in the way and it is sturdy enough to handle the load.
You might try to look at some jag suspensions to to get some ideas too.
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