Kedas Posted March 24, 2006 Share Posted March 24, 2006 Given data: 250Nm @ 1900 tr/min. (TDI 1.9L engine, VW Jetta 2006) 0-80 km/h (0-22.2m/s) in 8,0 sec Car weight: 1384Kg (empty) adding a person and fuel lets say 1500kg We can assume that the car speed is rising more or less linear between 0 and 80km/h. That is even logical if an automatic gear is trying to keep the tr/min around 1900tr/min resulting in a constant torque generated by the engine. The calculation: The average acceleration of the car is measured at 22.2 / 8 = 2.8 m/s² The needed average force to get this acceleration is 1500kg*2.8m/s² = 4200Newton (m*a=F) The torque is 4200N*r=250Nm --> r = 0.06meter (6cm) So what went wrong what does this 6cm mean? It isn't even remotely close to the tire radius although this 4200N should be present on the tires (2100N each) if you want an acceleration of 2.8m/s². Link to comment Share on other sites More sharing options...
Klaynos Posted March 24, 2006 Share Posted March 24, 2006 this is just a guess, but the torque is per driven wheel perhaps? That'd make r = 12cm which is more beliveable... Link to comment Share on other sites More sharing options...
Kedas Posted March 24, 2006 Author Share Posted March 24, 2006 We are looking for a torque of 4200N*0.3m=1260Nm Specifing the torque of an engine per wheel would be odd but that would get us close, 4*250Nm or 1000Nm. Link to comment Share on other sites More sharing options...
Milken Posted March 24, 2006 Share Posted March 24, 2006 Big guess because I do not remember 100% how torque is expressed. The first line of given data is measured /min. Data elsewhere is measured /s, converting the first line to seconds makes it smaller, resulting in a larger figure for the answer. Hopefully I haven't come across as a complete idiot. Link to comment Share on other sites More sharing options...
Kedas Posted March 24, 2006 Author Share Posted March 24, 2006 Big guess because I do not remember 100% how torque is expressed. The first line of given data is measured /min. Data elsewhere is measured /s, converting the first line to seconds makes it smaller, resulting in a larger figure for the answer. Hopefully I haven't come across as a complete idiot. The engine rotation speed of 1900tr/min isn't used in the calculation. It is assumed to be more or less constant resulting in a more or less constant torque of 250Nm. The torque varies a little with the rotation speed. But you have won one point for trying Link to comment Share on other sites More sharing options...
Norman Albers Posted March 24, 2006 Share Posted March 24, 2006 Ouch! Don't transmission ratios matter as we go? If an engine torques at an RPM, we then go through trans and differential where we trade torque for rotation speed? (Any lever trades force and distance.) This is fun and here is a corallary riddle: exactly where is a car propelled forward? On what pieces of what? Link to comment Share on other sites More sharing options...
Kedas Posted March 24, 2006 Author Share Posted March 24, 2006 Torque doesn't change with an higher gear but the ratio decreases resulting in a different force on the wheels for every gear, the higher the gear the lower the force on the wheels. (lower acceleration) A higher gear is the same like putting a bigger wheel on your car. Torque is the same but an increasing radius reduces the force. (Torque/radius=force) 6 gear system (fixed ratio:1/4.2) 1st, 1/3.42, total 1/14.35 (3.42*4.2) 2nd, 1/2.11, total 1/8.84 3rd, 1/1.43, total 1/6.00 4th, 1/1.09, total 1/4.57 5th, 1/1.10 total 1/3.64 6th, 1/0.91, total 1/3.02 ('total' is the ratio between the engine and the wheel) Ratio diff. between first and second is 1.6 so the force drops with a factor 1.6 when it's put in second gear. (assuming a constant torque of the engine) So the assumption of a more or less constant acceleration is wrong. We would need to known when they changed gear to compare engine torque with this 0-80km/h timing of 8sec. Link to comment Share on other sites More sharing options...
Norman Albers Posted March 24, 2006 Share Posted March 24, 2006 Torque at the engine is not changing but torque at the wheels is. Power is the product of torque and angular speed. That comes through unchanged except for loss. Link to comment Share on other sites More sharing options...
Milken Posted March 25, 2006 Share Posted March 25, 2006 Are the calculations wrong? Or are we to explain the answer? Link to comment Share on other sites More sharing options...
Norman Albers Posted March 25, 2006 Share Posted March 25, 2006 "A different force on the wheel" is a different torque at same. Link to comment Share on other sites More sharing options...
Kedas Posted March 25, 2006 Author Share Posted March 25, 2006 If I'm not mistaken this are the forces that pull the car. (without losses and at 1900tr/min 250Nm) 1st, 1/3.42, total 1/14.35, 250Nm*14.35=3587Nm on the tire, /0.3 = 12000N 2nd, 1/2.11, total 1/8.84, 250Nm*8.84=2210Nm on the tire, /0.3 = 7367N 3rd, 1/1.43, total 1/6.00, 250Nm*6=1500Nm on the tire, /0.3 = 5000N 4th, 1/1.09, total 1/4.57, 250Nm*4.57=1143Nm on the tire, /0.3 = 3810N 5th, 1/1.10 total 1/3.64, 250Nm*3.64=910Nm on the tire, /0.3 = 3033N 6th, 1/0.91, total 1/3.02, 250Nm*3.02=755Nm on the tire, /0.3 = 2516N Link to comment Share on other sites More sharing options...
c00ler Posted March 31, 2006 Share Posted March 31, 2006 Hi everyone sorry that bother you here. I'm landlord of online game http://www.f1grandprixmanager.com And i have an idea to create free online game based on car physics For this moment i have a simple model that run on the track and i can create graphs (rpm by distance, speed by distance and so on) Problem is that my knowledge of physics is not enough to create more powerfull model that will work with aerodrags and other things. I'm good in programming (that ability help to built http://www.f1grandprixmanager.com), but so good in physics. So if you have some free time to help with it i will be more than happy. If you interested i can give a link of my implementation of current physic model (just do not want to post it on online forum) Regards With Best Regards Max Rozanoff Link to comment Share on other sites More sharing options...
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