Swapping Detroit for CAT or Cummins

It doesn't take an engineering degree to understand basic gearing and how horsepower works

 

Splain to me hot rod

 

I do not have an engineering degree, thankfully, but I have ridden and do ride bicycles.
Over all ratio is all that matters. If your tire tread speed and your crank speed are the same between a 2.70 and a 4.10, your transmission ratio makes up the difference and other than parasitic losses within the trans, it takes the same power to over come the same hill.
Just like bicycles, if your ring tooth count and your cog tooth count are the same, it does not matter if you are in the big ring up front and big cog in the back or small ring and small cog. The ratio over all is what matters.
Again, I do not have a degree in either engineering or math, nor physics, nor........ I do have my own experience, and that really only matters to me. But if you can prove me wrong, I will listen and learn. I'm not going to let ego get in my way, I truly do try to learn and appreciate it greatly when people do teach me new things.
Now, if you wish to discuss applied torque to components in the above examples, then that is a whole different story. 2.70s will blow out drive shafts when down in the low trans gears applying high torque loads.

 

Alright, Agnes (I can use nicknames too), I’ll “splain” it to you and explain it to everyone else.

Horsepower is simply RPM multiplied by torque divided by 5252. An engine making 1400tq at 1500rpm is generating 400hp at the flywheel. Let's ignore parasitic loss for now and run that through an Eaton overdrive transmission with a .73 top gear and you get roughly 2054.8 rpm, 1022.39tq and 400hp when it hits the drive shaft. Now we have a final rearend ratio of 3.70. Run those numbers through and you get 555.5rpm, 3782 tq and 400hp at the tires. Notice a pattern with that hp number yet?


Take that same engine and swap the transmission to a direct drive and the rears to 2.70. The 1500rpm, 1400tq and 400hp are the same at the flywheel and at the drive shaft because it's straight in and straight out. But when you divide your rpm by 2.7 and multiply your torque by 2.7, you get 555.5 and 3782 respectively. That comes out to 400hp at the wheels.


The engine rpm, tq and hp are same. At the wheels, the rpm is the same, the torque is the same and the horsepower is the same on both trucks so why would one pull better on hills?


Please note that some numbers were rounded for simplicity sake.

 

Tell me why you can't get the 2.67 ratio in a high horse high torque applications

 

Because, like you, there are too many people who don't understand so they don't get them. If there's no demands why would a company make them?

 

Try again

 

Wow, you really can't fix stupid

 

Careful who you call stupid, I will make you out to be

 

Go for it. Start by explaining why one truck will pull better if both are putting the same numbers to the ground like in my example