Recent content by RedRook

Another way of finding torque at the wheel at top speed is looking at power of the forces resisting acceleration. I have to get to work, so I'll just say that the equations for A and C that you posted in the other thread are exactly what I think is correct. You can replace CRR with a more...

My post will be short today as time has just disappeared somehow. I'll go back and try to do some metric unit examples later. SI are what I'm used to, so it is easier for me. Today I think I'll just give some idea of where I'm heading with all of this in case someone wants to work it out faster...

Ok, you might be right about the gear box. I couldn't find the spot on the Renault site that gave the transmission details, so I looked on other sites. I went back to those sites, and I saw they also mentioned a petrol engine. They are obviously talking about a different car. When I finally got...

Ok, the problem here is the engine power drops off if you start adding a drop in voltage. This is the steep drop off point I mentioned. I'm aware of this point, but this point does not equal constant power. Your torque is dropping at that point, because your power is dropping much faster than...

What you are describing doesn't make sense according to the laws of physics. We know speed is going up, because you say that the car is going from 25mph to 70mph. We know the Flunce Z.E. doesn't have a continuously variable transmission, and you are saying you want to assume it is in one gear...

You know I take that back. Power can be constant on a CVT. A CVT actually constantly changes the gear ratio's to keep the engine at the same speed as the tires change speed. Whether they set it up for RPM's at peak torque or peak power will depend on if they care more about fuel economy or...

Most of your assumptions are wrong here. I don't mean to mean, they just are. For one, the engine torque does not equal the tire torque. If we are in top gear, then the tires normally have half the torque then the engine, because they are turning twice as fast. The power at the engine is also...

Correct. Static friction is a ratio that represents how well two things are stuck together. It is all that applies until we have movement between the objects. Once we have movement you kind of get a skipping motion on the microscopic level. Think of it like two pieces of metal that have fitting...

There are a couple of things to consider here. The first thing is understanding the forces on the object. F_M = \mu_s * F_N F_M = Maximum force of friction \mu_s = Static coefficient of friction F_N = Normal force Static friction doesn't always produce a force. It only produces a...

In an ideal world the power from the engine would go directly toward moving the vehicle. We already saw this was not happening in the engine. Gasoline has a given potential energy per mass, but the power of our engine did not increase purely based on the amount of gasoline burned. A four stroke...

So my starting point is the equation for acceleration based on power. A(t) = P(V(t))/ (M * V(t)) A = Acceleration t = Time P = Power V = Velocity M = Mass My first goal is to come up with an equation for power based on velocity as velocity changes over time. The first component of power that...

I played around in another thread with mathematically determining where a car should be time wise on 1/4 mile run. I had a chance to drive this car myself giving me more information than you normally have on a car you have yet to buy. As I thought more deeply about my calculations, I became...

I think your biggest mistake is probably where you started using T / r with T = engine torque, and r = wheel radius. The wheel radius alone is not enough, because the torque on the wheel is dependent both on the gear ratio and the velocity of the car. Unless you are assuming that power is on a...

What values are you using with units. It is hard to check where you might be going wrong, if I don't know the details of the object you are trying to make calculations for.

Ok, let me try this again. I want to be able to enter the appropriate V0 every run, so I need to work that into the equation. v2 = 2c2 [t/c −C1] With t = 0, I should have: v2 = -2c2C1 Meaning: C1 = v2/-2c2 Plugging that back into the equation gives us: v2 = 2c2 [t/c − v2/-2c2] = 2tc +...