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Old 10-07-2010, 09:22 AM   #16
supershaft
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Quote:
Originally Posted by Rob Farmer
You won't let it drop will you The only reason I fitted the early airbox on my PD was because the bike came with a mismatched engine, loads of dyno runs showed I had a problem with airflow at high rpm, the early airbox fixed this. When I get round to it with the exception of high comp pistons the engine will be put back to standard including the airbox. As a point of interest one of the guys over here spoke to HPN about a similar issue and was told the standard flat airbox doesn't have enough volume for the 40mm carbs, apparently if you make it 2cm deeper it makes a big difference to performance on 40mm carbed bikes. Morespeed also make a high volume replacement airbox and he's getting over 100 BHP out of his engines so there's something in it.


I have enough parts to put a complete 81 - 84 UK spec RS motor together, I also have the 78 squish band barrels and pistons that I haven't really got a use for. The whole point of the thread was to try and figure if there was any benefit to using them. I'll stick with the standard 9.5:1 pistons.

Good info though and thanks for the input. It is appreciated
I was just teasing about the airbox. Nevertheless, saying that the square airbox doesn't flow enough for 40mm Bings makes no sense. For instance, the most air an engine pulls is usually right at or around peak torque. Air flow and torque are very closely related. You can tune an engine with a square airbox to make a lot more torque and hp for that matter than any '77RS ever had. BTDT. No, I don't think that square airbox is the best thing on the planet but it is by far not the worse.

Increasing airbox volume? If I had a vintage road racer and a dyno, I think a LOT of power could be had by making a large still airbox. Even GP bikes have had airboxs for some time now! I think of them as being something like an air battery. Instead of your engine having to suck in the whole world every time an intake valve opens, it only has to worry about what is in the box. Of course, that box needs to refill VERY quickly but . . . . That's how I look at that forest.

Thanks for calling it good input. It's fun. Hopefully it can help someone. It's great to share experiences and opinions.
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Old 10-09-2010, 06:31 PM   #17
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Quote:
Originally Posted by supershaft
For instance, the most air an engine pulls is usually right at or around peak torque. Air flow and torque are very closely related.
No. Peak volumetric efficiency and peak torque generally coincide. Therefore, peak airflow per revolution and peak torque are closely related.

Airflow is proportional to volumetric efficiency and rpm. Provided everything is jetted right, peak airflow will be near peak power.

FWIW, I've never bought into the round air filter is better, though. It would be worth testing, but I haven't even found the time to go redyno my GS with the SR-Racing exhaust system.
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Old 10-09-2010, 07:59 PM   #18
supershaft
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Quote:
Originally Posted by 2ndlaw
No. Peak volumetric efficiency and peak torque generally coincide. Therefore, peak airflow per revolution and peak torque are closely related.

Airflow is proportional to volumetric efficiency and rpm. Provided everything is jetted right, peak airflow will be near peak power.

FWIW, I've never bought into the round air filter is better, though. It would be worth testing, but I haven't even found the time to go redyno my GS with the SR-Racing exhaust system.
You have lost me 2ndlaw. Don't you mean peak airflow PER INTAKE STROKE. When else does air flow? Does it somehow magically keep flowing all that time the intake valve is closed? What does rpm have to do with how efficient the intake stroke is? The highest volumetric efficiency IS the most air flow. Air flows one intake stroke at a time.

We agree that peak torque, peak volumetric efficiency, and peak airflow all occur about the same time per intake stroke or "revolution". I will agree that peak airflow occurs at peak power like you said if you agree that peak torque is peak power. After all, hp is just a figment of our imagination. Surely you understand that peak torque is peak power for any engine that does not rev above 5252 rpm. HP is ALWAYS less than torque below that rpm. Peak hp and peak torque are always at exactly the same rpm below 5252rpm. Above that magical rpm, hp is always greater than torque but it is still absolutely tied to torque production and thusly airflow. Hp will continue to rise after peak torque as long as the revs are picking up faster than the torque is falling off. Peak hp isn't peak power, it's peak amounts of power in a given time frame. Us humans have made that time frame up. Peak power happened back at peak torque along with peak volumetric efficiency. In other words, when peak hp is above peak torque, it is because the engine can manage to suck less air faster for a while. How are you figuring peak air flow and peak volumetric efficiency are different?

supershaft screwed with this post 10-09-2010 at 08:06 PM
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Old 10-11-2010, 10:31 AM   #19
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Quote:
Originally Posted by supershaft
You have lost me 2ndlaw. Don't you mean peak airflow PER INTAKE STROKE. When else does air flow? Does it somehow magically keep flowing all that time the intake valve is closed? What does rpm have to do with how efficient the intake stroke is? The highest volumetric efficiency IS the most air flow. Air flows one intake stroke at a time.SNIP How are you figuring peak air flow and peak volumetric efficiency are different?
The highest airflow per stroke (unit air charge) is at the highest volumetric efficiency.

Some math:
Unit air charge=volumetric efficiency*displacement*density of air
Airflow/time=unit air charge at a given speed*speed (speed meaning rpm, rps, etc.)
intake strokes/time=engine speed/2 (for a four stroke)
Airflow/time=speed/2*volumetric efficiency*displacement*air density

Peak torque is around where airflow/stroke is maximized.
Peak power is around where airflow/time is maximized.

Here are some quotes from Obert's "IC Engines and Air Pollution",1973. Obert was probably the best engineer ever as far as explaining things in a practical matter. Much better than me!

"For a hypothetical SI or CI engine at WOT with constant percentage heat loss, the indicated torque and the unit air charge are directly proportional at each engine speed...Practically, it can be considered that the point of maximum torque occurs at essentially the same speed as that for maximum air charge."

"Suppose that the unit air charge at wot (in lb per stroke) were to be multiplied by the number of strokes per hour of the engine. The result would be the air consumption of the engine (My note: what we call "air flow" here.) (in lb per hour)...Here it can be realized that that both speed and unit air charge control air consumption and that the air consumption continues to climb even though the unit air charge has passed its maximum point. This increase is finally halted by a rapid fall in the unit air charge...For a hypothetical engine with constant percentage heat loss, the indicated horsepower and air consumption are directly proportional."

Hope that helps.
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Old 10-11-2010, 10:43 AM   #20
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Quote:
Originally Posted by supershaft
After all, hp is just a figment of our imagination. Surely you understand that peak torque is peak power for any engine that does not rev above 5252 rpm. HP is ALWAYS less than torque below that rpm. Peak hp and peak torque are always at exactly the same rpm below 5252rpm. Above that magical rpm, hp is always greater than torque but it is still absolutely tied to torque production and thusly airflow.
Horsepower is not a figment of our imagination. It is one unit for power, the rate at which useful work can be accomplished.

Torque is a measure of force.

In other words, torque let's us know if we can climb a hill, power let's us know how fast.

Power=torque*angular speed. If speed increases faster than torque falls with increasing speed, power increases. 5252 has nothing to do with it.
The 5252 is just a unit conversion factor when torque is in ft-lbs and power is in horsepower. It has no other physical significance.
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Old 10-11-2010, 11:15 AM   #21
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Quote:
Originally Posted by 2ndlaw
The highest airflow per stroke (unit air charge) is at the highest volumetric efficiency.

Some math:
Unit air charge=volumetric efficiency*displacement*density of air
Airflow/time=unit air charge at a given speed*speed (speed meaning rpm, rps, etc.)
intake strokes/time=engine speed/2 (for a four stroke)
Airflow/time=speed/2*volumetric efficiency*displacement*air density

Peak torque is around where airflow/stroke is maximized.
Peak power is around where airflow/time is maximized.

Here are some quotes from Obert's "IC Engines and Air Pollution",1973. Obert was probably the best engineer ever as far as explaining things in a practical matter. Much better than me!

"For a hypothetical SI or CI engine at WOT with constant percentage heat loss, the indicated torque and the unit air charge are directly proportional at each engine speed...Practically, it can be considered that the point of maximum torque occurs at essentially the same speed as that for maximum air charge."

"Suppose that the unit air charge at wot (in lb per stroke) were to be multiplied by the number of strokes per hour of the engine. The result would be the air consumption of the engine (My note: what we call "air flow" here.) (in lb per hour)...Here it can be realized that that both speed and unit air charge control air consumption and that the air consumption continues to climb even though the unit air charge has passed its maximum point. This increase is finally halted by a rapid fall in the unit air charge...For a hypothetical engine with constant percentage heat loss, the indicated horsepower and air consumption are directly proportional."

Hope that helps.
Now you are making sense. In that last quoted sentence, Obert forgets to mention his whole point: air consumption and hp OVER A PERIOD OF TIME are directly proportional. That has been my whole point as well. That doesn't change the fact that the most air an engine draws in is at peak torque. With no time frame peak torque IS peak power. Any torque peak limited to below 5252 rpm results in the peak torque and peak hp being at exactly the same rpm regardless. A lot of engines don't rev over 5000 rpm. More air is more air. More air over time is more air over time. I thought it was clear that I was talking about more air. Now you are talking about more air AND more air over time. That's fine as long as you understand the difference.

I am hoping this conversation helps or sparks an interest in understanding.
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Old 10-11-2010, 12:12 PM   #22
supershaft
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Quote:
Originally Posted by 2ndlaw
Horsepower is not a figment of our imagination. It is one unit for power, the rate at which useful work can be accomplished.

Torque is a measure of force.

In other words, torque let's us know if we can climb a hill, power let's us know how fast.

Power=torque*angular speed. If speed increases faster than torque falls with increasing speed, power increases. 5252 has nothing to do with it.
The 5252 is just a unit conversion factor when torque is in ft-lbs and power is in horsepower. It has no other physical significance.
A mile has nothing to do with hp? Hp is a calculation and 5252 is at its center. 5252 has no significance at all unless you want to measure and talk about hp. And like Obert, there you go talking about power again within a time frame without mentioning the time frame. Speed is relative.

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Old 10-11-2010, 12:47 PM   #23
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Originally Posted by wirewrkr
The pistons themselves are not designated to the material of the cylinders, it's the rings that are different between cast iron and nikasil.
Not true.
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Old 10-11-2010, 05:17 PM   #24
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Quote:
Originally Posted by supershaft
Now you are making sense. In that last quoted sentence, Obert forgets to mention his whole point: air consumption and hp OVER A PERIOD OF TIME are directly proportional. That has been my whole point as well. That doesn't change the fact that the most air an engine draws in is at peak torque. With no time frame peak torque IS peak power. Any torque peak limited to below 5252 rpm results in the peak torque and peak hp being at exactly the same rpm regardless. A lot of engines don't rev over 5000 rpm. More air is more air. More air over time is more air over time. I thought it was clear that I was talking about more air. Now you are talking about more air AND more air over time. That's fine as long as you understand the difference.

I am hoping this conversation helps or sparks an interest in understanding.
1. The pressure drop through components such as an air filter is a function of air flow per unit time. If we say something doesn't have enough airflow for, e.g., 40's, it really means that the pressure drop through that component becomes great enough at these larger airflows (BTW, just saying flow means per unit time) to effect performance. (There are also nonsteady compressible effects but the math becomes very complicated.

2. The statement that if an engine doesn't turn over 5000 rpm peak torque and peak horsepower must be at the same rpm is completely false. Go here for some examples from flatheads: http://www.vanpeltsales.com/FH_web/f...cs-85early.htm
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Old 10-11-2010, 05:54 PM   #25
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Quote:
Originally Posted by supershaft
A mile has nothing to do with hp? Hp is a calculation and 5252 is at its center. 5252 has no significance at all unless you want to measure and talk about hp. And like Obert, there you go talking about power again within a time frame without mentioning the time frame. Speed is relative.
No. Horsepower is a unit of power. Power is the rate at which work can be done.

In general, P=T*N, where P is power, T is torque and N is angular speed. The number that concerns you, 5252, comes from the result of multiplying T is ft-lbf by N in RPM, which has results of ft*lbf*rev/min. We need a conversion,

(2*PI*rad/rev)*(min/60 sec)*(1 HP/ (550 *ft*lbf/sec)

We need rad(ians) because that's the way it works in physics, and 1 HP = 550*ft*lbf/sec.

Multiplying that all out gives you your 1/5252. Miles have nothing to do with it.

Or, in metric:

T (in N*m)*N (in rev/min)*(2*PI*rad/rev)*(1 min/60 sec) gives you P in (Watts). Or, in the familiar form:

P (in Watts)=T (in N*m) * RPM/9.55
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Old 10-11-2010, 06:58 PM   #26
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Not true.

Explain.
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Old 10-11-2010, 10:11 PM   #27
supershaft
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Quote:
Originally Posted by 2ndlaw
No. Horsepower is a unit of power. Power is the rate at which work can be done.

In general, P=T*N, where P is power, T is torque and N is angular speed. The number that concerns you, 5252, comes from the result of multiplying T is ft-lbf by N in RPM, which has results of ft*lbf*rev/min. We need a conversion,

(2*PI*rad/rev)*(min/60 sec)*(1 HP/ (550 *ft*lbf/sec)

We need rad(ians) because that's the way it works in physics, and 1 HP = 550*ft*lbf/sec.

Multiplying that all out gives you your 1/5252. Miles have nothing to do with it.

Or, in metric:

T (in N*m)*N (in rev/min)*(2*PI*rad/rev)*(1 min/60 sec) gives you P in (Watts). Or, in the familiar form:

P (in Watts)=T (in N*m) * RPM/9.55
Thanks, I don't know where I came up with the mile from. Nevertheless, 5252 has everything to do with dyno charts. And, more to the point, when peak hp is at a higher rpm than peak torque, it is because the engine can manage to suck less air faster. It's a classic case where less is more when you factor in time.

It reminds me of when a 220 hp engine doesn't make enough power to replace a 160hp engine. The 220 engine is cranking out 220hp at 2100 rpm and the 160hp engine is cranking out 160hp at 1100 rpm. The 160hp engine is cranking out 200ftlbs more torque than the 220hp engine. Guess which one is sucking more air and making more power?
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Old 10-12-2010, 03:31 PM   #28
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Quote:
Originally Posted by 2ndlaw
1. The pressure drop through components such as an air filter is a function of air flow per unit time. If we say something doesn't have enough airflow for, e.g., 40's, it really means that the pressure drop through that component becomes great enough at these larger airflows (BTW, just saying flow means per unit time) to effect performance. (There are also nonsteady compressible effects but the math becomes very complicated.

2. The statement that if an engine doesn't turn over 5000 rpm peak torque and peak horsepower must be at the same rpm is completely false. Go here for some examples from flatheads: http://www.vanpeltsales.com/FH_web/f...cs-85early.htm
Who says that just saying flow means per unit of time? You still haven't explained how an engine draws more air than peak volumetric efficiency when you don't consider a unit of time? Why would they call it peak volumetric efficiency if something else is really peak volumetric efficiency? Is it because we have been using the word 'flow' instead of 'draw'.

Why has every other dyno chart I have ever seen have torque and hp converge at 5252 rpm? On every chart I have ever seen up until now the peak hp and peak torque are always at the same rpm until the engine revs over 5252rpm. It has to be or the graphs won't converge there. What gives?
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Old 10-12-2010, 09:13 PM   #29
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Quote:
Originally Posted by 2ndlaw
1. The pressure drop through components such as an air filter is a function of air flow per unit time. If we say something doesn't have enough airflow for, e.g., 40's, it really means that the pressure drop through that component becomes great enough at these larger airflows (BTW, just saying flow means per unit time) to effect performance. (There are also nonsteady compressible effects but the math becomes very complicated.

2. The statement that if an engine doesn't turn over 5000 rpm peak torque and peak horsepower must be at the same rpm is completely false. Go here for some examples from flatheads: http://www.vanpeltsales.com/FH_web/f...cs-85early.htm
Now I had time to check out those weird dyno charts. What I said still holds water. Check out those charts. The left and right side have different values. I have never seen charts like those before. They are like two completely different graphs in one. What's the point? If both the torque curves and the hp curves had the same gradients on the graph you would see that I am correct. I didn't catch that at first either. Very strange!
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Old 10-13-2010, 07:19 PM   #30
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Originally Posted by lkchris
Not true.
Yeah, Explain.
Cuz that's wrong.
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