I have been wondering just how much torque actually goes to each individual mount.

Lets say you have 500ft/lbs of torque. Is this an exertion of 500ft/lbs on each engine mount or is 500ft/lbs cummulative of what opposing force the engine mounts are putting against the engine individually (250ft/lbs each)?

Obviously this is ignoring the fact that your differential is a torque multiplier even if the torque is measured on a 1:1 basis within the transmission such as 5th gear.

 

a quick assumption would be 250 ft-lb toque applied to the drivers side mount, with the forces pointing in an upward direction, pulling up on the mount. on the passenger side, 250ft-lb with the forces aimed downward through the mount.

of course, this isn't taking account the static load of the weight of the engine either.

It has been about 15 years since I took statics and dymanics as pasrt of a BSME, so my assumtions may not be correct, and/or I'm forgetting something.

 

Thanks for the response.

Correct me if I am wrong here, but if a car makes that much torque on a Dyno to the wheels would the engine be putting the wheel measurement worth on the mounts? Is it less due to the torque multipliers it goes through to get to the wheel to give it that 500ft/lbs?

 

The only way you will get the full tourqe to the mounts is if the tires don't turn. by the time you reach 500 pounds your down the highway....I would estimate you get about 80 pounds in a quick burst on slicks.

 

The torque at the wheels is not what is actually reported by a dyno as it corrects for speed, which is gear dependent. A better way of saying this is that most dynos today report "torque of the engine as measured at the wheels". That is why most dyno operators do not have to correct for a rear final drive gear swap.

In regards to the torque of the motor being applied to the engine - I believe that is an oversimplification because the pressure against the heads are balanced in a V configuration as the cylinders fire and the drivetrain loss stemming from friction of the rotating assembly is the only thing that favors counter-rotation to driveshaft spin.


EDIT: I have a different engine mount question (that I hope the OP won't mind me interjecting). If you install solid engine mounts, it is widely held that this reduces drivetrain loss. My belief is that this is not the case, and in fact the drivetrain loss is still equivalent, but results in more heat, rather than heat and movement from the force of friction (of oil shearing against the engine body). Just speculation though - any thoughts?

 

also, dont forget the tranny mount taking some load off of the front motor mounts. this is the stuff you will just go crazy thinking about

 

It reduces the drivetrain loss, tourqe is an energy like electricity, it will find the path of least resistance. If you didn't have mounts at all ( and somehow levitated the engine) it would spin under the hood rather than turn the wheels (engine weighs less than the car)

I think it is such a small amount of loss I wouldn't bother unless you don't mind a less smooth ride and happen to be changing the motor.

 

I don't think this is quite true because the force applied to the heads is balanced other than as noted in my prior post. Even if the wheels locked - the amount of rotational torque on the engine mounts would not be equivalent to the full engine output (as the cylinder forces would counteract each other without a spinning assembly).

 

It would be equivalent to the number measured at the flywheel.

 

Not that it really matters, but for academic discussion...

I don't think that's correct because the number measured at the flywheel is the accumulated torque of 6 cylinders, thanks to the rotating crank. At the engine mounts, each bank would offset each other, reducing the net force seen at any given time.