Log's interpretation required (not college course)

[Probability]

I have a graph produced by a electronics software program that shows the rate of increase of injection pressure over a period of 160 seconds while the engine runs up to 4000 RPM.

The fuel pressure is shown up the y axis and is in log form, now the confusion for me seems to be that log e is 2.718..., but the graph log scale shows log E, so am assuming because I can't find any other information to tell me otherwise that log E is base 10!

Now the problem. I know that common rail injection systems pressures run at around an operating pressure of 1800 bar, or there about.

With the graph comes a table of data, so re-producing one line of it here, it reads; 1.7999e+006.

I think I can work this out as 1.7999 [second function] log which is 10^006 = 1799900

Now the graph shows hectopascals hPa but the table of data shows log e!

I also know that there are 1000 hPa = 1 bar

So if I am correct, then;

1799900 / 1000 = approx 1800 bar.

I can do this all day long from the table of data and it reads correct, but from the graph I am struggling because the y axis log's are written as;

2E + 09, 4E + 09, 6E + 09, 8E + 10, 10E + 10 .....

These logs are not, or do not appear the same as the logs in the table of data, so when I try to work out the hPa from say 1.8E + 10, it does not relate to 1800 bar, yet;

from the table of data I can achieve;

1.8E^006 = 1800000 and when divided by 1000

1800000 / 1000 = 1800 bar.

Can anyone advise if I need a different method to work out the hPa pressure using 1.8E+10 methods please!

 

[stapel]

I have a graph produced by a electronics software program that shows the rate of increase of injection pressure over a period of 160 seconds while the engine runs up to 4000 RPM.

The fuel pressure is shown up the y axis and is in log form, now the confusion for me seems to be that log e is 2.718...,

No; the natural exponential, the number "e", is equal to about 2.718. The base-e log of e, being ln(e) (and being pronounced as "the natural log of e"), is equal to 1. Any base log, when taking the log of its own base, returns a value of 1.

...but the graph log scale shows log E, so am assuming because I can't find any other information to tell me otherwise that log E is base 10!

Why?

Now the problem. I know that common rail injection systems pressures run at around an operating pressure of 1800 bar, or there about.

With the graph comes a table of data, so re-producing one line of it here, it reads; 1.7999e+006.

I think I can work this out as 1.7999 [second function] log which is 10^006 = 1799900

Does this match what is on the graph?

Now the graph shows hectopascals hPa but the table of data shows log e!

Since, as discussed above, "ln(e)" is just "1", I doubt that this is what is being shown. Would it be possible to provide pictures of what you're seeing?

I also know that there are 1000 hPa = 1 bar

What is a "bar"?

So if I am correct, then;

1799900 / 1000 = approx 1800 bar.

I can do this all day long from the table of data and it reads correct, but from the graph I am struggling because the y axis log's are written as;

2E + 09, 4E + 09, 6E + 09, 8E + 10, 10E + 10 .....

These logs are not, or do not appear the same as the logs in the table of data, so when I try to work out the hPa from say 1.8E + 10, it does not relate to 1800 bar, yet;

from the table of data I can achieve;

1.8E^006 = 1800000 and when divided by 1000

1800000 / 1000 = 1800 bar.

Can anyone advise if I need a different method to work out the hPa pressure using 1.8E+10 methods please!

When you reply, please explain how you feel that the graph you're trying to interpret requires the techniques of differential equations. Thank you!

 

[Probability]

No; the natural exponential, the number "e", is equal to about 2.718. The base-e log of e, being ln(e) (and being pronounced as "the natural log of e"), is equal to 1. Any base log, when taking the log of its own base, returns a value of 1.


Why?


Does this match what is on the graph?


Since, as discussed above, "ln(e)" is just "1", I doubt that this is what is being shown. Would it be possible to provide pictures of what you're seeing?


What is a "bar"?


When you reply, please explain how you feel that the graph you're trying to interpret requires the techniques of differential equations. Thank you!

Without getting too heavily involved, the post is in the wrong section, sorry my fault. The third law of logarithms is what I was looking for along with a clear understanding of the difference between e and E.

A bar is not what you lean on in the local pub drowning your sorrows after a bad day

The designer has used logs on the graph and shown them as 1.8E+10, the capital E was the confusion, thus not knowing what base it represented. Now I know,

n log A = log A^n

1.8 log_10 = log_10^10 = 1.8_10^10

That is a long number, which is why the designer chose the log scale to represent the data.