Given the engine supplys the power...how is the wheel torque higher than the drive shaftRx7man said:Yes, you just need to know all the gear ratios in between. There will be some loss from friction of course...
So if your rear wheel torque is 1000 lb-ft of torque, and your rear axle ratio is 4.10:1, you will have 1000/4.1 = 244 lb-ft toque in the driveshaft... Repeat for the transmission gear ratio and you'll have engine torque
The torque is higher because of the gear reduction in the differential... POWER output would be slightly less because of inefficiencies in the differential thoughHybrid_engine said:Given the engine supplys the power...how is the wheel torque higher than the drive shaft
So if that is true. Which i agree it is true. What would be the effect of having an engine that operates like a wheel that spins like a rotary engine where the leverage is increase by 500% at the same rpm? Leverage being 5 times longer. Does 5 times more leverage equal 5 times the power At the same rpm?Rx7man said:The torque is higher because of the gear reduction in the differential... POWER output would be slightly less because of inefficiencies in the differential though
Rx7man said:I don't quite follow what you're asking... If you could increase the stroke 5 times longer So the offset is 5 times the other engines main to rod journal distance. Would you have 5 times the pressure on the crankshaft?
If you had intake 93% the size of the piston diameter with turbos that had pressure to get 100% volumetric efficiency at 1000 rpm to introduce fresh air at top of cylinder and allow exit of combustion charge out exhaust ports at BDC area where it takes place from 120 degrees to 210 degrees. But you only compressed air from 330 degress ATDC to TDC and only compressed bore and stroke equal distance. 10 inch bore and only compressed 10 inch of stroke. it free wheeled "no compression until 330 degrees ATDC to TDC with 5 times the friction only as added drag on rotation. Would it have 5 times the torque? And what would be the effect if the engine had no negative G forces from any parts changing trajectory "as in having parts going up and down? Let's say every moving part was in an orbit? With only centripetal and centrifugal force with stored energy from inertia "mass in an orbit tied to a central point by centripetal force?Rx7man said:Yes, you would, but increasing the stroke by 5x usually involves a very serious redesign of an engine.. If you keep the same bore size, you're going to have 5x the displacement, but lose a lot of volumetric efficiency due to very small valves, and the peak RPM of the engine will drop by about the same ratio as your increased stroke.. you'll make more torque, but it's unlikely you'll make much more POWER.. (the valves can only flow so much air)
If you keep to a near 'square' engine design (bore ~= stroke), and increase the stroke and bore 5x, you'll have 125x (5x5x5) the displacement, at that point, yes, you'll definitely make more power and torque of course, but the thing will be MASSIVE
Not talking about either one of those designs. Something very different. No valves yet it let's air in and out and compresses from 330 degrees to TDC. Combustion from 350 degrees to after TDC. it has 10 inch pistons running on diesel. Can you calculate the pressure on the rotation given 78 sq. in. on piston with very standard diesel pressure curve in the cylinder? Keep in mind the expansion is 40 inches before it hits exhaust.Rx7man said:You're losing me again.. Intake PORT or intake VALVE? Is the exhaust a port or a valve?
If you're talking about ported engines like most 2 cycle engines, they'll pretty much always be symmetrical.. The port that opens 120* ATDC will close 120* BTDC, so it's impossible to get 120* of power stroke with 30* of compression stroke
If you're talking about a cam/valve timed engine, it's impossible to get an intake valve 93% of the piston diameter...
The offset is 25 inch from centerline. As compared to a 25 inch offset of rod journal to main centerHybrid_engine said:Not talking about either one of those designs. Something very different. No valves yet it let's air in and out and compresses from 330 degrees to TDC. Combustion from 350 degrees to after TDC. it has 10 inch pistons running on diesel. Can you calculate the pressure on the rotation given 78 sq. in. on piston with very standard diesel pressure curve in the cylinder? Keep in mind the expansion is 40 inches before it hits exhaust.
http://www.thecartech.com/subjects/engine/engine_formulas.htmHybrid_engine said:The offset is 25 inch from centerline. As compared to a 25 inch offset of rod journal to main center
Why don't you start your own topic and describe in detail what you're trying to accomplish.. At this point we're chasing our tailsHybrid_engine said:
Wheel torque is the amount of rotational force applied to the wheels of a vehicle, while engine torque is the amount of rotational force produced by the engine. Wheel torque is affected by factors such as gear ratios and friction, while engine torque is a measure of the engine's power output.
To calculate engine torque from wheel torque, you need to know the gear ratio of the vehicle and the radius of the wheels. You can then use the equation: Engine torque = Wheel torque x Gear ratio x Wheel radius.
Working back from wheel torque to engine torque allows us to understand the power and performance of the engine. It also helps in troubleshooting and diagnosing issues related to the engine or transmission.
Yes, it is possible for wheel torque to be greater than engine torque. This can happen when the vehicle is in a lower gear, which increases the gear ratio and multiplies the engine torque. Additionally, external factors such as road conditions and vehicle weight can also affect wheel torque.
Wheel torque can significantly impact a vehicle's handling. If the wheel torque is too high, it can cause the wheels to spin and lose traction, leading to a loss of control. On the other hand, too little wheel torque can result in a lack of power and responsiveness, making it difficult to control the vehicle.