# How much force is needed to push a piston down? [closed]

Combustion from fuel and air is used to push a piston down. When they combustion it creates a lot of force, my question is how much force is needed to push the piston all the way down when the vehicle is turned on and moving in many different surfaces?

How much force will then be needed?

Is it relatively easy to push down or do it require a great deal of force?

• If I have the head off of a motor many times I can push it down with my hand. If the engine is in gear and on a flat surface it would take considerably more. Try and wrap some parameters around your question so you get a more defined and quality answer. Cheers! – DucatiKiller May 5 '16 at 0:32
• I really don't know that this is answerable, mainly because there are many factors involved in figuring this out. Some engine builders, when looking for that last ounce of HP/TQ out of their engine, will measure the drag the piston with rings on it has in the cylinder. The rings have a huge affect on how much drag is on the pistons. Then when you put the connecting rod on a crankshaft, there's another factor. Then when you tie the rest of the pistons/con-rods onto the crank, there's another factor. It all piles onto the amount of force it takes and it all depends on how the engine is built. – Pᴀᴜʟsᴛᴇʀ2 May 5 '16 at 0:36
• But are there any numbers that would be to low? So for example, would 50 pounds on the piston be to little on average? – LostPecti May 5 '16 at 0:38
• @Pᴀᴜʟsᴛᴇʀ2 makes a good point. On a 1987 Kawasaki 750 the crank horsepower is 101. Horsepower at the rear wheel is 86. That's a 14% power loss. If you scaled that 200HP it would still have about the same power loss but obviously the percentage of loss would be lower. That should give you some idea of how much power is lost from the piston to the rear wheel. – DucatiKiller May 5 '16 at 0:39
• @DucatiKiller - Actually, power loss would be about the same percentage, everything but the power being equal. For instance, you can figure an automatic car would lose about 18-20% at 400 crank HP, but would still lose 18-20% at 600 crank HP. The loss is linear. – Pᴀᴜʟsᴛᴇʀ2 May 5 '16 at 0:44

I apologize if my question wasn't as technical as you want. I would like to add I have no knowledge of cars and I'm just learning. I thought it was common sense to ask questions if you didn't know something. I am sure at one point you had no knowledge either. Its not a hard understand that no one knows anything until they sit down and learn. So do you understand? Good ;) – LostPecti

No problem. I'm all for it. But the question really isn't that simple. Its not just a simple push, its a push over a particular time, When your piston is at perfect top dead center or at perfect bottom center, no amount of pushing move the piston. You could add tons of force, but the piston just won't move. To begin to understand the details, and the why questions, you will have to invest some time in your own education starting at the basics. I didn't know a thing about cars till I started reading a Haynes type service manuals and car magazines. 3 years later I was an automotive engineer in Detroit. I get it. I really do.

But all is not lost. There are tools available to help you understand some of this stuff better. I'd start at Kahn Academy, and sign up for the physics classes. The Kahn Academy classes are totally free. You'll want "Work and Energy", "Thermodynamics", "Impacts and Linear Momentum", "Newtons laws of motion" and "Moments / torque" That should get you the ability to accurately answer your own question.

What I'm saying here, is its not really productive to attempt to summarize a whole lot of information on how something works in a simple internet forum question. There is no right answer to your question in the way its asked.

The biggest thing to remember is that the "push" isn't really a simple push. Its a controlled chemical reaction (explosion) inside a chamber timed to go off with maximum force at a particular time in the crankshaft rotation cycle. And that force isn't just felt at a single piston position. It varies as the crankshaft turns. The explosion START is controlled via spark plug timing. Fuel / air mix and exhaust timing is based on the camshaft which opens and closes valves.

There is no way to accurately answer your question. If you want to know more, folks here will happily guide you in your education. I'm certainly available (my email is available via a simple click on my ID) Not trying to make you feel bad. Again, no offense is intended.

But if you really want to know the answer to your question (and how to make your car go faster :^) you will have to invest a bit of time in your own education on the topic. I'm a strong believer in making your hobby your career.

Update:

I was looking for a diagram that showed gasoline engine pressure vs. crank rotation angle. I found this powerpoint presentation. I thought it pretty sound. The question about pressure to push a piston down is only part of the conversation. Don't forget you have to use that push to get the piston to make one more spin so it will exhaust the burnt air/fuel out of the cylinder. The pressure chart is on slide #18, and you can see that the max PRESSURE is around 20 degrees after Top Dead Center (TDC). In that chart the ignition spark starts before TDC (funny I knew that but never really thought about it till today...) It takes a bit of time for the air/fuel/spark explosion to propagate.

To get the force on the piston (at a particular engine speed and load) you'd have to integrate the pressure from TDC to Bottom Dead Center AND account for the crankshaft angle (via sine/cosine relationship) in your integral. Its not a simple problem...

Edit, link to powerpoint presentation `Internal Combustion Engines, Valve Timing Diagrams, Ujjwal K. Saha, Phd`, updated. Ouch. Apparently that link has a short lived authentication token attached to it. Do a search look for the paper here...

• Great non answer. +1 – DucatiKiller May 5 '16 at 7:05
• Thank you for your explanation. I really appreciate your help in guiding me in the right direction. I will make sure I begin my studies first thing in the morning as it is late now. Really thank you. And no offense has been taken. – LostPecti May 5 '16 at 7:20