03-09-2015, 09:39 AM
Gentlemen,
In a related thread there are a myriad of attempted explanations about horsepower and torque. Some of the responses are on the mark, most of them are not. Our moderator Mickey asked some excellent questions and I have done my best answer them here. I am copying my post from that thread in post #3 here as a prelude to answering Mick's questions.
_____________________________
What I don't understand is if there is this symbiotic relationship between torque and HP expressed in the formula - Torque x RPM = HP, why don't the lines on dyno charts reflect this relationship and mirror each other?
Why would the lines be any different?
Why does it not show a corresponding dip in horsepower where it shows a dip in torque?
Why would there be a dip in torque when rpm is increasing? One chart is going up in the same spot the other chart is going down and vice versa.
Do you see what I am saying?
As far as carbs and cams and valves go, they are the same ones in the motor when you start the dyno runs as when you end the dyno runs, and they are made of metal so they don't radically alter shape (lift duration etc.) from 2000-6000 rpms where the dip in torque shows up on the dyno chart of 600s I posted.
I will probably never understand this so the explanation is moot I suppose. It just doesn't make any sense to me. With my last breaths I will probably still be saying torque makes you go and HP makes you go fast. Lol
________________________________
Mickey,
I’m going to start by answering your specific questions and then I’ll go into why.
---“What I don’t understand is if there is this symbiotic relationship between torque and horsepower expressed in the formula - Torque X RPM = horsepower, why don’t the lines on dyno charts reflect this relationship and mirror each other?” –
A - The answer to this question is because torque is constantly changing, both up AND down, as RPM increases. The reason why torque would decrease as engine speed increases I will explain later, there are many reasons for this, and the two curves (torque and horsepower) cannot mirror each other because the horsepower curve involves a second variable and that is RPM whereas the torque curve does not involve engine RPM. Stay with me and this will become clear as well.
---“ Why does it not show a corresponding dip in horsepower where it shows a dip in torque?”--
A - Because the torque number is multiplied by RPM to arrive at horsepower, as long as torque is decreasing at a slower rate than RPM is increasing, horsepower will continue to rise. So if torque decreases by 10% with an increase in engine speed of 20%, horsepower will continue to increase even though torque has decreased. Again you are asking yourself, “why would torque decrease as engine speed increases?”. Again, I will explain that below so just take what I say here as a given for now.
--- “Why would there be a dip in torque when rpm is increasing?--
A - In an internal combustion engine, Torque is the force applied to the top of the piston when the air fuel mixture is ignited, pushing that piston straight down. In other words, how hard is each “individual” push on the top of the piston? The force of each individual push changes primarily because of two different variables. 1. How much fuel/air mixture is in that cylinder at the time the spark plug lights it off and 2. How optimum is that mixture? The optimum mixture of fuel and air in a modern internal combustion engine is approximately 13 to 1, or 13 parts air for every 1 part fuel. As you deviate from that optimum mixture power will decrease. At a mixture of 10 to 1 each explosion will produce less power than a mixture of 13 to 1. A mixture of 15 to 1 will also produce less power than that optimum 13 to 1 mixture. Another component also comes into play and that is exhaust gas contamination which takes up room in the cylinder yet produces no power. So maximum torque is achieved when we get the maximum amount of air/fuel mixture into the cylinder, with as little contamination as possible, and at the optimum ratio.
A good way to visualize torque is to imagine you are driving a long nail into a block of wood with a hammer. As you swing the hammer down upon that nail and strike it, it penetrates into the wood. Obviously if you strike the nail harder it will penetrate farther with each blow. The force of each blow of that hammer is torque. So will the strongest carpenter always drive the nail in fastest? After all, he hits the nail harder than any other carpenter. The answer is no and the reason is that we need to know how many times he strikes that nail each minute. If the strong carpenter drives the nail .25 inches deep with each blow but only swings the hammer four times a minute, it will take him a full minute to drive the 1 inch long nail in. If a weaker carpenter is only able to hit that nail hard enough to drive it .10 inches deep with each blow but instead of four strikes a minute he hits the nail with 20 strikes a minute, the weaker carpenter will have that same nail driven all the way in, in just 30 seconds. The weaker carpenter who is applying less torque during each hammer blow has double the horsepower of the stronger carpenter because he is operating at an RPM of 20 strikes a minute where the stronger carpenter with more torque is only operating at an RPM of four strikes a minute. In this example the weaker carpenter has twice the horsepower of the stronger carpenter.
Torque is the force of each individual blow of that hammer or, in an engine, the force of each individual push upon a piston.
Horsepower is a measure of the amount of work that an engine, or a carpenter, can accomplish and it is expressed by the formula - torque X RPM = horsepower. Is it starting to make sense now?
Which brings us to the question of “Why would torque decrease when engine RPM increases?” After all, as Mickey has accurately stated, “As far as carbs and cams and valves go, they are the same ones in the motor when you start the dyno runs as when you end the dyno runs, and they are made of metal so they don't radically alter shape (lift duration etc.) from 2000-6000 rpms where the dip in torque shows up on the dyno chart of 600s I posted.”
A – Mick is right in his statement above, but there are other factors that do not remain constant and those other factors are the reason that torque can fall off, or dip, in the middle of an engine’s RPM range and why torque ALWAYS falls off rapidly when engine speed increases near an engines redline speed. First we must understand that an engine is nothing more than an air pump. Any gas will move from an area of high pressure to an area of lower pressure and this simple principle is what makes an engine work. Here I want to use another example to help you visualize what is going on. I assume most of us have held a medical syringe in our hands at some point and, with the needle on, you could note that when you pulled the plunger out there was a lot of resistance because the air could not flow into the body of that syringe very fast through the tiny opening in that needle. As such, when you pulled the plunger out you were drawing a partial vacuum and it might have taken a couple seconds for enough air to flow in through that tiny opening before the syringe was completely filled with air. If you took the needle off that syringe which would leave a much larger opening for air to flow through you could still pull the plunger out fast enough to draw a partial vacuum but now instead of a couple seconds that were required to fill that syringe with the restriction of a needle, it might only take one half of a second before the body of the syringe was completely filled. In both cases, the faster you pulled the plunger out to create that vacuum, the faster the body of the syringe started to fill BUT…. If you pulled the plunger out twice as fast, would that syringe fill with air twice as fast? The answer is no. Even if you pulled that plunger out extremely fast creating a nearly complete vacuum inside the body of the syringe, the atmospheric pressure on the other side (14.7 pounds per square inch at sea level) is the only pressure available to push air through and the restriction of the small needle opening will only allow so much air to flow per second. By taking the needle off the air will flow much faster through the larger orifice however we run into the same limitations. Once we pulled that plunger fast enough to create vacuum quicker than air pressure allows that gas to flow through the orifice allowing it into the cylinder, a harder pull ceases to fill that syringe faster.
Now let’s introduce another variable that we are dealing with in an internal combustion engine and that is time. Obviously, if our syringe was an engine cylinder and the plunger in that syringe was the piston in an engine, the cylinder would eventually fill completely even through valves as small as the hole in that syringe IF there was enough time for air to flow through that extremely restrictive intake. But that’s not how an engine works. Go back to our syringe example and imagine that with the needle on, we pulled the plunger out rapidly and then after 1 second we put our finger over the opening in the top of the needle stopping airflow into the syringe. The syringe would not be completely filled because that restrictive opening requires two seconds to flow enough air to fill the syringe. Do the same thing with that syringe with the needle removed and it would be completely filled because it only takes .5 seconds to flow enough air through the now less restrictive opening. But if we really speed things up and choke off those openings after only .25 seconds, neither syringe would be completely filled. The syringe with the needle would only be 1/8 filled and the syringe without the needle would be 1/2 filled.
And this is what happens when an internal combustion engine nears it’s redline. An intake valve opens and closes faster and faster as the engine turns faster and faster. No matter how big the intake valves are they are still restrictive and when the “duration” of their opening gets short enough the cylinders are no longer completely filling with air/fuel mixture. When an engine is turning 8000 RPM the intake valves are only open for one half the time that they are open when that same engine is turning 4000 RPM. As long as the time available allows for the complete filling of that cylinder it doesn’t matter much that the intake valve opening duration is getting shorter. But at some point, like our syringe example, the intake valve opening duration get’s so short that there just isn’t enough time for air to flow through and completely fill that cylinder so we are now lighting off the spark plugs with lesser and lesser charges of fuel/air mixture. When this lesser charge explodes the “hammer blow” is not pushing on the piston as hard as each individual push was at a lower RPM which allowed for a complete cylinder fill.
At first, these less powerful hammer blows are made up for by their increasing frequency as RPM increases. The carpenter is hitting the nail with slightly less force but the nail is still going in faster because he is hitting the nail more times per minute. And so horsepower continues to go up even though torque is going down because RPM is multiplying the smaller hits faster than the force of those hits is decreasing. But soon as intake valve duration gets so short that cylinder fills are really falling off, torque decreases at a faster rate than RPM increases and horsepower starts to fall off rapidly as well. There are other limitations to engine speed because the stress on engine components goes up to the square of RPM so airflow is not the only limitation but for the sake of this discussion I’m leaving these other factors out for now.
---“Why would torque decrease in the middle of an engines RPM range and then increase again?”
A - A number of factors could cause this. With the knowledge that our engine is an air pump and knowing that gases move primarily because of pressure differentials, we will acknowledge now a second component to the movement of gases through an orifice or a passageway and that second component is momentum or velocity. Pressure differential can be created one of two ways, by increasing pressure on one side or decreasing pressure on the other side. An example of this would be moving air through a garden hose by placing your lips on one end and blowing or going to the other end of the hose and sucking. Obviously, a combination of the two would move the most air through that hose. In an engine, when a piston moves down in the cylinder it is sucking, or creating a vacuum which allows ambient air pressure (14.7 pounds per square inch at sea level) to rush in and fill that vacuum. To move even more air into that cylinder you could blow into the intake using either exhaust driven turbochargers or belt driven superchargers. Such devices can double or even triple the amount of fuel/air mixture forced into the cylinder during each cycle, producing a lot more power. An example of the smaller component of momentum would be turning one end of the hose so that the opening faced into the wind and then sucking on the other end. Here we are using a little bit of ambient velocity to aid in our creation of pressure differential. In this way we can move slightly more air than we could if there were no wind, or air velocity (momentum). Two examples of using this principle on vehicles would be the hood scoop on a car or the velocity stacks you see sticking up out of the top of old hot rod engines. The tuned intakes on most modern motorcycles utilize air velocity in this manner to produce slightly more power as well.
As an internal combustion engine operates there are times when both the intake valves and the exhaust valves are open at the same time. This is called valve overlap. When the exhaust valves open and the high-pressure air starts to rush out of the cylinder, the momentum of that escaping charge helps draw additional vacuum because it’s mass and motion does not stop when pressure inside the cylinder drops to zero. Much the same as a ball does not stop moving after you stop pushing it across the floor. It continues to roll for a while. Depending upon valve overlap, total valve area, and several other factors, engines all have sweet spots where escaping exhaust is exiting smoothly helping to draw additional incoming fresh fuel/air mixture during that valve overlap leaving minimal spent exhaust gases to contaminate the fresh air/fuel charge. At a slightly higher or lower engine speed turbulence may build up impeding this optimum flow resulting in more contamination or a fuel air mixture that isn’t quite optimum. This can result in less powerful hammer blows in that RPM range but again, as long as RPM is increasing at a higher percentage than torque is decreasing, horsepower will continue to increase even in the range of that dip in torque.
This is been a rather lengthy explanation both for you to read, and for me to write. If it’s helped at all please consider it a thank you to Mick for all the time and effort that he puts in here on behalf of all of us. All the best.
Chip
In a related thread there are a myriad of attempted explanations about horsepower and torque. Some of the responses are on the mark, most of them are not. Our moderator Mickey asked some excellent questions and I have done my best answer them here. I am copying my post from that thread in post #3 here as a prelude to answering Mick's questions.
_____________________________
What I don't understand is if there is this symbiotic relationship between torque and HP expressed in the formula - Torque x RPM = HP, why don't the lines on dyno charts reflect this relationship and mirror each other?
Why would the lines be any different?
Why does it not show a corresponding dip in horsepower where it shows a dip in torque?
Why would there be a dip in torque when rpm is increasing? One chart is going up in the same spot the other chart is going down and vice versa.
Do you see what I am saying?
As far as carbs and cams and valves go, they are the same ones in the motor when you start the dyno runs as when you end the dyno runs, and they are made of metal so they don't radically alter shape (lift duration etc.) from 2000-6000 rpms where the dip in torque shows up on the dyno chart of 600s I posted.
I will probably never understand this so the explanation is moot I suppose. It just doesn't make any sense to me. With my last breaths I will probably still be saying torque makes you go and HP makes you go fast. Lol
________________________________
Mickey,
I’m going to start by answering your specific questions and then I’ll go into why.
---“What I don’t understand is if there is this symbiotic relationship between torque and horsepower expressed in the formula - Torque X RPM = horsepower, why don’t the lines on dyno charts reflect this relationship and mirror each other?” –
A - The answer to this question is because torque is constantly changing, both up AND down, as RPM increases. The reason why torque would decrease as engine speed increases I will explain later, there are many reasons for this, and the two curves (torque and horsepower) cannot mirror each other because the horsepower curve involves a second variable and that is RPM whereas the torque curve does not involve engine RPM. Stay with me and this will become clear as well.
---“ Why does it not show a corresponding dip in horsepower where it shows a dip in torque?”--
A - Because the torque number is multiplied by RPM to arrive at horsepower, as long as torque is decreasing at a slower rate than RPM is increasing, horsepower will continue to rise. So if torque decreases by 10% with an increase in engine speed of 20%, horsepower will continue to increase even though torque has decreased. Again you are asking yourself, “why would torque decrease as engine speed increases?”. Again, I will explain that below so just take what I say here as a given for now.
--- “Why would there be a dip in torque when rpm is increasing?--
A - In an internal combustion engine, Torque is the force applied to the top of the piston when the air fuel mixture is ignited, pushing that piston straight down. In other words, how hard is each “individual” push on the top of the piston? The force of each individual push changes primarily because of two different variables. 1. How much fuel/air mixture is in that cylinder at the time the spark plug lights it off and 2. How optimum is that mixture? The optimum mixture of fuel and air in a modern internal combustion engine is approximately 13 to 1, or 13 parts air for every 1 part fuel. As you deviate from that optimum mixture power will decrease. At a mixture of 10 to 1 each explosion will produce less power than a mixture of 13 to 1. A mixture of 15 to 1 will also produce less power than that optimum 13 to 1 mixture. Another component also comes into play and that is exhaust gas contamination which takes up room in the cylinder yet produces no power. So maximum torque is achieved when we get the maximum amount of air/fuel mixture into the cylinder, with as little contamination as possible, and at the optimum ratio.
A good way to visualize torque is to imagine you are driving a long nail into a block of wood with a hammer. As you swing the hammer down upon that nail and strike it, it penetrates into the wood. Obviously if you strike the nail harder it will penetrate farther with each blow. The force of each blow of that hammer is torque. So will the strongest carpenter always drive the nail in fastest? After all, he hits the nail harder than any other carpenter. The answer is no and the reason is that we need to know how many times he strikes that nail each minute. If the strong carpenter drives the nail .25 inches deep with each blow but only swings the hammer four times a minute, it will take him a full minute to drive the 1 inch long nail in. If a weaker carpenter is only able to hit that nail hard enough to drive it .10 inches deep with each blow but instead of four strikes a minute he hits the nail with 20 strikes a minute, the weaker carpenter will have that same nail driven all the way in, in just 30 seconds. The weaker carpenter who is applying less torque during each hammer blow has double the horsepower of the stronger carpenter because he is operating at an RPM of 20 strikes a minute where the stronger carpenter with more torque is only operating at an RPM of four strikes a minute. In this example the weaker carpenter has twice the horsepower of the stronger carpenter.
Torque is the force of each individual blow of that hammer or, in an engine, the force of each individual push upon a piston.
Horsepower is a measure of the amount of work that an engine, or a carpenter, can accomplish and it is expressed by the formula - torque X RPM = horsepower. Is it starting to make sense now?
Which brings us to the question of “Why would torque decrease when engine RPM increases?” After all, as Mickey has accurately stated, “As far as carbs and cams and valves go, they are the same ones in the motor when you start the dyno runs as when you end the dyno runs, and they are made of metal so they don't radically alter shape (lift duration etc.) from 2000-6000 rpms where the dip in torque shows up on the dyno chart of 600s I posted.”
A – Mick is right in his statement above, but there are other factors that do not remain constant and those other factors are the reason that torque can fall off, or dip, in the middle of an engine’s RPM range and why torque ALWAYS falls off rapidly when engine speed increases near an engines redline speed. First we must understand that an engine is nothing more than an air pump. Any gas will move from an area of high pressure to an area of lower pressure and this simple principle is what makes an engine work. Here I want to use another example to help you visualize what is going on. I assume most of us have held a medical syringe in our hands at some point and, with the needle on, you could note that when you pulled the plunger out there was a lot of resistance because the air could not flow into the body of that syringe very fast through the tiny opening in that needle. As such, when you pulled the plunger out you were drawing a partial vacuum and it might have taken a couple seconds for enough air to flow in through that tiny opening before the syringe was completely filled with air. If you took the needle off that syringe which would leave a much larger opening for air to flow through you could still pull the plunger out fast enough to draw a partial vacuum but now instead of a couple seconds that were required to fill that syringe with the restriction of a needle, it might only take one half of a second before the body of the syringe was completely filled. In both cases, the faster you pulled the plunger out to create that vacuum, the faster the body of the syringe started to fill BUT…. If you pulled the plunger out twice as fast, would that syringe fill with air twice as fast? The answer is no. Even if you pulled that plunger out extremely fast creating a nearly complete vacuum inside the body of the syringe, the atmospheric pressure on the other side (14.7 pounds per square inch at sea level) is the only pressure available to push air through and the restriction of the small needle opening will only allow so much air to flow per second. By taking the needle off the air will flow much faster through the larger orifice however we run into the same limitations. Once we pulled that plunger fast enough to create vacuum quicker than air pressure allows that gas to flow through the orifice allowing it into the cylinder, a harder pull ceases to fill that syringe faster.
Now let’s introduce another variable that we are dealing with in an internal combustion engine and that is time. Obviously, if our syringe was an engine cylinder and the plunger in that syringe was the piston in an engine, the cylinder would eventually fill completely even through valves as small as the hole in that syringe IF there was enough time for air to flow through that extremely restrictive intake. But that’s not how an engine works. Go back to our syringe example and imagine that with the needle on, we pulled the plunger out rapidly and then after 1 second we put our finger over the opening in the top of the needle stopping airflow into the syringe. The syringe would not be completely filled because that restrictive opening requires two seconds to flow enough air to fill the syringe. Do the same thing with that syringe with the needle removed and it would be completely filled because it only takes .5 seconds to flow enough air through the now less restrictive opening. But if we really speed things up and choke off those openings after only .25 seconds, neither syringe would be completely filled. The syringe with the needle would only be 1/8 filled and the syringe without the needle would be 1/2 filled.
And this is what happens when an internal combustion engine nears it’s redline. An intake valve opens and closes faster and faster as the engine turns faster and faster. No matter how big the intake valves are they are still restrictive and when the “duration” of their opening gets short enough the cylinders are no longer completely filling with air/fuel mixture. When an engine is turning 8000 RPM the intake valves are only open for one half the time that they are open when that same engine is turning 4000 RPM. As long as the time available allows for the complete filling of that cylinder it doesn’t matter much that the intake valve opening duration is getting shorter. But at some point, like our syringe example, the intake valve opening duration get’s so short that there just isn’t enough time for air to flow through and completely fill that cylinder so we are now lighting off the spark plugs with lesser and lesser charges of fuel/air mixture. When this lesser charge explodes the “hammer blow” is not pushing on the piston as hard as each individual push was at a lower RPM which allowed for a complete cylinder fill.
At first, these less powerful hammer blows are made up for by their increasing frequency as RPM increases. The carpenter is hitting the nail with slightly less force but the nail is still going in faster because he is hitting the nail more times per minute. And so horsepower continues to go up even though torque is going down because RPM is multiplying the smaller hits faster than the force of those hits is decreasing. But soon as intake valve duration gets so short that cylinder fills are really falling off, torque decreases at a faster rate than RPM increases and horsepower starts to fall off rapidly as well. There are other limitations to engine speed because the stress on engine components goes up to the square of RPM so airflow is not the only limitation but for the sake of this discussion I’m leaving these other factors out for now.
---“Why would torque decrease in the middle of an engines RPM range and then increase again?”
A - A number of factors could cause this. With the knowledge that our engine is an air pump and knowing that gases move primarily because of pressure differentials, we will acknowledge now a second component to the movement of gases through an orifice or a passageway and that second component is momentum or velocity. Pressure differential can be created one of two ways, by increasing pressure on one side or decreasing pressure on the other side. An example of this would be moving air through a garden hose by placing your lips on one end and blowing or going to the other end of the hose and sucking. Obviously, a combination of the two would move the most air through that hose. In an engine, when a piston moves down in the cylinder it is sucking, or creating a vacuum which allows ambient air pressure (14.7 pounds per square inch at sea level) to rush in and fill that vacuum. To move even more air into that cylinder you could blow into the intake using either exhaust driven turbochargers or belt driven superchargers. Such devices can double or even triple the amount of fuel/air mixture forced into the cylinder during each cycle, producing a lot more power. An example of the smaller component of momentum would be turning one end of the hose so that the opening faced into the wind and then sucking on the other end. Here we are using a little bit of ambient velocity to aid in our creation of pressure differential. In this way we can move slightly more air than we could if there were no wind, or air velocity (momentum). Two examples of using this principle on vehicles would be the hood scoop on a car or the velocity stacks you see sticking up out of the top of old hot rod engines. The tuned intakes on most modern motorcycles utilize air velocity in this manner to produce slightly more power as well.
As an internal combustion engine operates there are times when both the intake valves and the exhaust valves are open at the same time. This is called valve overlap. When the exhaust valves open and the high-pressure air starts to rush out of the cylinder, the momentum of that escaping charge helps draw additional vacuum because it’s mass and motion does not stop when pressure inside the cylinder drops to zero. Much the same as a ball does not stop moving after you stop pushing it across the floor. It continues to roll for a while. Depending upon valve overlap, total valve area, and several other factors, engines all have sweet spots where escaping exhaust is exiting smoothly helping to draw additional incoming fresh fuel/air mixture during that valve overlap leaving minimal spent exhaust gases to contaminate the fresh air/fuel charge. At a slightly higher or lower engine speed turbulence may build up impeding this optimum flow resulting in more contamination or a fuel air mixture that isn’t quite optimum. This can result in less powerful hammer blows in that RPM range but again, as long as RPM is increasing at a higher percentage than torque is decreasing, horsepower will continue to increase even in the range of that dip in torque.
This is been a rather lengthy explanation both for you to read, and for me to write. If it’s helped at all please consider it a thank you to Mick for all the time and effort that he puts in here on behalf of all of us. All the best.
Chip
