How do i find the equation of this graph?

[shawnz1102]

Hi everyone, I have a graph here (not a homework related, just something i encountered in daily life) but it's in increments of X=0.25 and Y=25 I'd like to change mimic this graph and change the increments to X=0.5 and Y=25 but I'm not sure what the equation is...

For all the math experts here, does anyone know how to find the equation just by looking at the graph? It's a "dyno sheet" of a car.

Thanks in advance!

 

[HallsofIvy]

What graph? I see four different graphs here. Theoretically, if really knew the graph, in the sense of knowing what "y" is for every possible value of "x", you could find the "equation" but there is no general method- and there is no guarantee that the equation of a given graph can be written in terms of "elementary" functions or, for that matter, in terms of functions that anyone has ever defined!

In practice, the best you can do is fit a curve to a finite number of points- and there exist an infinite number of functions passing through a finite number of given points. For example, there always exist an n-1 degree polynomial passing through any given n points.

 

[shawnz1102]

You totally lost me there with the mathematical terms haha.

Now that i thought about it, there isn't really an equation to be defined... The graph is basically a graphical plot of different data. I guess I could find the X and Y at every single grid line and draw a dot for them, and then connecting the dots as accurately as possible using the graph as reference. What do you think?

Edit: This is a horsepower vs torque graph, and I'm trying to find the distance between the horsepower curve and the torque curve. I was thinking that maybe there is a relation or ratio between the two, and I could use that ratio and multiply it by the horsepower to find the torque or something.

 

[UndeniablyRex]

Horsepower is a measure of, oddly enough, energy per time. Torque is a measure of force at a distance(from the axis of rotation). They can't be converted unless you know some other information about the engine you are investigating.

Also, what is the x-axis in your graph?

What HallsofIvy is saying is to get a few points off each curve(just estimate as accurately as possible to their points) and put them through a computer or calculator and get a regression.

 

[UndeniablyRex]

but it's in increments of X=0.25 and Y=25 I'd like to change mimic this graph and change the increments to X=0.5 and Y=25

I don't think you're saying what you want to. All you have to do to accomplish this is to erase every other line on the x-axis starting with .25. This would leave you with a line at 0, .5, 1, 1.5, etc... as you stated you wanted.

 

[Mark44]

Also, what is the x-axis in your graph?

The horizontal axis is RPM, with units in RPM X 1000. The graph is torque vs. HP for some engine run on a dynomometer.

I'm trying to find the distance between the horsepower curve and the torque curve. I was thinking that maybe there is a relation or ratio between the two, and I could use that ratio and multiply it by the horsepower to find the torque or something.

The distance between the two curves doesn't make any sense because the two graphs are measuring different things. For example, at 4000 RPM, the engine is putting out about 167 HP and 213 ft-lb or torque. You can certainly subtract 167 from 213, but what would be the units of the resulting difference?

One thing to notice is that the two curves (really two pairs of curves) cross at what looks like 5250 RPM. They actually cross at 5252 RPM, which they always do, as a result of the formula that engine tuners use to calculate HP from torque, which is what a dynomometer actually measures. The formula is HP = RPM/5252 * Torque, and maybe this is what you're looking for. At any given RPM, you can read off the torque value and calculate the HP value.

You can see from this formula that at 5252 RPM, HP and torque will be numerically equal. (The units are not right, though, as HP is ft-lb/sec while torque is in ft-lb.) This formula assumes that English units are used.

Out of curiousity, what type of engine are you running on the dyno? It looks to be a pushrod engine, due to the relatively low RPMs, and looks like it might be a small V8 or V6, or maybe even be a large displacement V Twin motorcycle engine.

 

[shawnz1102]

Horsepower is a measure of, oddly enough, energy per time. Torque is a measure of force at a distance(from the axis of rotation). They can't be converted unless you know some other information about the engine you are investigating.

Also, what is the x-axis in your graph?

What HallsofIvy is saying is to get a few points off each curve(just estimate as accurately as possible to their points) and put them through a computer or calculator and get a regression.

The X axis is RPM range.

I don't think you're saying what you want to. All you have to do to accomplish this is to erase every other line on the x-axis starting with .25. This would leave you with a line at 0, .5, 1, 1.5, etc... as you stated you wanted.

Right, but the thing is, I don't have the graph on my computer. That image is the only data I have.

 

[shawnz1102]

The horizontal axis is RPM, with units in RPM X 1000. The graph is torque vs. HP for some engine run on a dynomometer.


The distance between the two curves doesn't make any sense because the two graphs are measuring different things. For example, at 4000 RPM, the engine is putting out about 167 HP and 213 ft-lb or torque. You can certainly subtract 167 from 213, but what would be the units of the resulting difference?

One thing to notice is that the two curves (really two pairs of curves) cross at what looks like 5250 RPM. They actually cross at 5252 RPM, which they always do, as a result of the formula that engine tuners use to calculate HP from torque, which is what a dynomometer actually measures. The formula is HP = RPM/5252 * Torque, and maybe this is what you're looking for. At any given RPM, you can read off the torque value and calculate the HP value.

You can see from this formula that at 5252 RPM, HP and torque will be numerically equal. (The units are not right, though, as HP is ft-lb/sec while torque is in ft-lb.) This formula assumes that English units are used.

Out of curiousity, what type of engine are you running on the dyno? It looks to be a pushrod engine, due to the relatively low RPMs, and looks like it might be a small V8 or V6, or maybe even be a large displacement V Twin motorcycle engine.

Yes Mark, that's exactly what I'm looking for! , and you are correct, it is a V6. It's actually the engine off a 2007 Nissan 350z.

Regarding the formula: HP = RPM/5252 * Torque...
I would first need to have either the HP or Torque available in order to calculate one of the other. Here, I'll give you my purpose for what I want to do, and maybe you can better advice me on how to achieve it. Basically, I am adding a piping to my car, and that dyno sheet above is of my car before having the pipes. I'd like to make another dyno sheet to approximate my HP and torque gains of adding the new piping. The only information I have with me is that at about 2,500 RPM, there will be a torque increase of about 20 ft/tq over the dyno sheet above.

 

[Mark44]

I don't think there's a way of doing what you want analytically (i.e., without running the engine on the dyno again). Just knowing the torque increase at one data point won't help you determine the torque and HP increases over the entire RPM range.

There's a whole art to getting performance increases by changing to freer-flowing exhaust headers and pipes. You can increase torque and HP in some RPM ranges, but decrease them in other ranges, depending on whether exhaust scavenging at the exhaust ports is improved or lessened. Improving exhaust flow is only one of the bag of tricks engine tuners use, with some of the others being cam changes, head porting to improve flow, etc.