Do V8 swaps mess up weight distribution? A lesson in Physics.
#76
#77
Open up! Search Warrant!
Joined: May 2002
Posts: 3,838
Likes: 3
From: Kicking down doors in a neighborhood near you
I have a degree in applied science (Aerospace Maintenance Technology) and was an aircraft mechanic for 9 yrs in the military. One of my jobs was to determine the weight and balance of an aircraft along the mean aerodynamic chord (an imaginary line starting at the nose and going through the tail). The purpose was to keep the center of gravity in a location that will prevent the plane from tipping over, or having too much weight in a location that will adversely affect how it handles in flight. If too much weight was in the nose, it could be difficult for the plane to climb. Too much fuel in the left wing tank could cause problems with lateral motion.
To do this, you need the original weight of the components you are removing, and for the component you are replacing; in this case, a rotary engine and a piston engine. Next, you need the distance from the original center of gravity (or what ever point you are measuring as the fulcrum, as this is the fixed point that the weight is rotating around.) I'm not sure of the original CG of an RX-7, but I'm sure it can be found. The distance from the center to the point where the force is acting on the car is the moment arm (imagine a lever).
Basically, here is how to mathmatically figure the new CG. CG shift = (weight of object moved) * (distance moved) / (total weight)
Example below:
Model RC car = 48 oz
car battery = 4 oz
battery moved forward - 3inches
4oz* 3in /48oz = .25 inches. So if I move the battery forward 3 inches, my car CG moves forward .25 inches. Same for moving backwards, or sideways.
To swap an engine, you have to calculate the weight of object moved as the difference between the two objects. If you install a piston engine that weighs 50lbs more than a rotary, your weight of object moved is 50 lbs. Hope that kinda helps without being too technical or confusing.
To do this, you need the original weight of the components you are removing, and for the component you are replacing; in this case, a rotary engine and a piston engine. Next, you need the distance from the original center of gravity (or what ever point you are measuring as the fulcrum, as this is the fixed point that the weight is rotating around.) I'm not sure of the original CG of an RX-7, but I'm sure it can be found. The distance from the center to the point where the force is acting on the car is the moment arm (imagine a lever).
Basically, here is how to mathmatically figure the new CG. CG shift = (weight of object moved) * (distance moved) / (total weight)
Example below:
Model RC car = 48 oz
car battery = 4 oz
battery moved forward - 3inches
4oz* 3in /48oz = .25 inches. So if I move the battery forward 3 inches, my car CG moves forward .25 inches. Same for moving backwards, or sideways.
To swap an engine, you have to calculate the weight of object moved as the difference between the two objects. If you install a piston engine that weighs 50lbs more than a rotary, your weight of object moved is 50 lbs. Hope that kinda helps without being too technical or confusing.
#78
#79
I have a degree in applied science (Aerospace Maintenance Technology) and was an aircraft mechanic for 9 yrs in the military. One of my jobs was to determine the weight and balance of an aircraft along the mean aerodynamic chord (an imaginary line starting at the nose and going through the tail). The purpose was to keep the center of gravity in a location that will prevent the plane from tipping over, or having too much weight in a location that will adversely affect how it handles in flight. If too much weight was in the nose, it could be difficult for the plane to climb. Too much fuel in the left wing tank could cause problems with lateral motion.
To do this, you need the original weight of the components you are removing, and for the component you are replacing; in this case, a rotary engine and a piston engine. Next, you need the distance from the original center of gravity (or what ever point you are measuring as the fulcrum, as this is the fixed point that the weight is rotating around.) I'm not sure of the original CG of an RX-7, but I'm sure it can be found. The distance from the center to the point where the force is acting on the car is the moment arm (imagine a lever).
Basically, here is how to mathmatically figure the new CG. CG shift = (weight of object moved) * (distance moved) / (total weight)
Example below:
Model RC car = 48 oz
car battery = 4 oz
battery moved forward - 3inches
4oz* 3in /48oz = .25 inches. So if I move the battery forward 3 inches, my car CG moves forward .25 inches. Same for moving backwards, or sideways.
To swap an engine, you have to calculate the weight of object moved as the difference between the two objects. If you install a piston engine that weighs 50lbs more than a rotary, your weight of object moved is 50 lbs. Hope that kinda helps without being too technical or confusing.
To do this, you need the original weight of the components you are removing, and for the component you are replacing; in this case, a rotary engine and a piston engine. Next, you need the distance from the original center of gravity (or what ever point you are measuring as the fulcrum, as this is the fixed point that the weight is rotating around.) I'm not sure of the original CG of an RX-7, but I'm sure it can be found. The distance from the center to the point where the force is acting on the car is the moment arm (imagine a lever).
Basically, here is how to mathmatically figure the new CG. CG shift = (weight of object moved) * (distance moved) / (total weight)
Example below:
Model RC car = 48 oz
car battery = 4 oz
battery moved forward - 3inches
4oz* 3in /48oz = .25 inches. So if I move the battery forward 3 inches, my car CG moves forward .25 inches. Same for moving backwards, or sideways.
To swap an engine, you have to calculate the weight of object moved as the difference between the two objects. If you install a piston engine that weighs 50lbs more than a rotary, your weight of object moved is 50 lbs. Hope that kinda helps without being too technical or confusing.
#80
He's dealing with our equivelant of sprung wieght, not suspension. Planes operate IN a medium, cars operate ON a surface. All the things he worries about on an airplane when it comes to weight and balance transfer pretty directly to a car and its dynamics.
Cars have the suspension system to act as a crutch for our weight placement sins and at worst we'll overload a tire. The car won't be as fast/quick/stable as it could be. The airplane, if not balanced, will end up on the 6 o'clock news looking like a smoking crater in someones roof or yard.
Cars have the suspension system to act as a crutch for our weight placement sins and at worst we'll overload a tire. The car won't be as fast/quick/stable as it could be. The airplane, if not balanced, will end up on the 6 o'clock news looking like a smoking crater in someones roof or yard.
#83
as builder of race cars? lol
biggest difference from a car is that an airplane has (almost....) one 1 pivot point called the lift center. Static margin is based on the lift center and center of gravity. This point is mesured exactly by others engineer buddies. Moving this point a bit too much will make your plane uncontrollable, exactly like a fucked up suspension... Have balance transfert on a plane, lol. Dont hit the brake pedal too hard
BTW, planes are designed unstable, his job is to make sure it is not too unstable or worst, perfectly stable.
biggest difference from a car is that an airplane has (almost....) one 1 pivot point called the lift center. Static margin is based on the lift center and center of gravity. This point is mesured exactly by others engineer buddies. Moving this point a bit too much will make your plane uncontrollable, exactly like a fucked up suspension... Have balance transfert on a plane, lol. Dont hit the brake pedal too hard
BTW, planes are designed unstable, his job is to make sure it is not too unstable or worst, perfectly stable.
#84
If where the weight is doesn't matter on a plane, just the center of lift and COG, why can't I do aerobatics with fuel in the wings of an Extra300?
His point, and mine, is that where you put the weight matters and its effect can be given a value that can be compared with some math. On an airplane the problems will manifest themselves differently than a car.
Comparing and apple and an orange can be tough but the final test for an airplane is a trip to the scales. They don't fix a weight problem by jacking one wheel with a coilover shock. So your wrong about him dealing with suspension.
Here is something else for you, I havn't posted updates in a while but you'll get the ideahttps://www.rx7club.com/showthread.p...ght=eprod+born
His point, and mine, is that where you put the weight matters and its effect can be given a value that can be compared with some math. On an airplane the problems will manifest themselves differently than a car.
Comparing and apple and an orange can be tough but the final test for an airplane is a trip to the scales. They don't fix a weight problem by jacking one wheel with a coilover shock. So your wrong about him dealing with suspension.
Here is something else for you, I havn't posted updates in a while but you'll get the ideahttps://www.rx7club.com/showthread.p...ght=eprod+born
#85
Open up! Search Warrant!
Joined: May 2002
Posts: 3,838
Likes: 3
From: Kicking down doors in a neighborhood near you
Gizmo is right in the fact that the dynamics that work against a motor vehicle to cause it to exceed its performance envelope is different in a plane. In a car, the body movement around the suspension, and the suspension movement itself play big role in CG. The formula I provided is what you use to find the static CG. However, it can be used to find the static CG of a car, boat, computer desk, etc.
Like he said, its a Static CG, to calculate a starting point. But nonetheless, it is important because if you know how to calculate where your new starting point is with your new piston engine, you will be able to calculate all the other factors. If you click on the link below and look halfway down the page (the whole page is good reading) then you will see formula for calculating CG during body roll, weight transfer during lateral forces, weight vs downforce and CG and steering affects. The site is courtesy of FFR (Factory Five Racing)
http://images.google.com/imgres?imgu...3Doff%26sa%3DN
Like he said, its a Static CG, to calculate a starting point. But nonetheless, it is important because if you know how to calculate where your new starting point is with your new piston engine, you will be able to calculate all the other factors. If you click on the link below and look halfway down the page (the whole page is good reading) then you will see formula for calculating CG during body roll, weight transfer during lateral forces, weight vs downforce and CG and steering affects. The site is courtesy of FFR (Factory Five Racing)
http://images.google.com/imgres?imgu...3Doff%26sa%3DN
#86
I never said the dynamics where the same, I said the weight and where it is matters to both. I don't think Gizmo put it in those words either. I also think he read what I wrote wrong, it probably could have been worded better now that I've re-read it myself. The "weight and balance transfer pretty directly..." part. I was saying the info transfers, not the balance. I'm not sure I even know what 'balance transfer' is lol
#87
the center of gravity has nothing to do with bump steer. bump steer is a purely geometric thing based off of the way the suspension moves. The only possible connection a lowered CG will have with bump steer is taht a lowered CG will make the car roll less, giving less bound of the outside wheels and jounce of the insides, thus giving less bump-steer.
It is mathematically possible to make the center of gravity actually be below ground level, this is not possible on a solid axle of course, but any independently suspended end of the car. When this occurs your steering geometry does some weird and wonderful things one of which is bump steer. The CG definately can contribute to bump steer, as can other factors influencing the angle of the front geometry.
The ideal center of gravity is about 1" above ground level, this has nothing to do with the weight of the car.
The last sentence is extremely simplistic, explain how this works with a solid rear axle. You can lower the CG on a solid rear by the way. oh and by the way are you saying that lowering a car's center of gravity doesn't affect the geometry, this would be just wrong.
Steve
#89
Originally Posted by aussiesmg
I don't think you actually understand what bump steer is, the center of gravity is a mathematic point which the car rotates laterally around, there is one at the rear and one at the front between the point where the tires actually contact the ground.
It is mathematically possible to make the center of gravity actually be below ground level, this is not possible on a solid axle of course, but any independently suspended end of the car. When this occurs your steering geometry does some weird and wonderful things one of which is bump steer. The CG definately can contribute to bump steer, as can other factors influencing the angle of the front geometry.
The ideal center of gravity is about 1" above ground level, this has nothing to do with the weight of the car.
The last sentence is extremely simplistic, explain how this works with a solid rear axle. You can lower the CG on a solid rear by the way. oh and by the way are you saying that lowering a car's center of gravity doesn't affect the geometry, this would be just wrong.
Steve
It is mathematically possible to make the center of gravity actually be below ground level, this is not possible on a solid axle of course, but any independently suspended end of the car. When this occurs your steering geometry does some weird and wonderful things one of which is bump steer. The CG definately can contribute to bump steer, as can other factors influencing the angle of the front geometry.
The ideal center of gravity is about 1" above ground level, this has nothing to do with the weight of the car.
The last sentence is extremely simplistic, explain how this works with a solid rear axle. You can lower the CG on a solid rear by the way. oh and by the way are you saying that lowering a car's center of gravity doesn't affect the geometry, this would be just wrong.
Steve
howi
#90
Amazing statements
Some of the people posting here should at least finish high school before they try to impress us with their knowledge of physics. And anyone who says V8 powered RX-7s are useless as road cars is really showing his ignorance. I have owned a lot of sports cars, including early Triumphs, Alfas, MGs, to E Type Jag, Porsches and a variety of Japanese cars. My Ford powered FB is a very nice road car, though of course it will never corner like a late model Vette. But it does corner better than my rotary powered FB because of the larger wheels and tires.
I believe the constant emphasis on 50-50 weight distribution is a lot of nonsense. The installation of a reasonable V8 will not significantly affect the weight distribution, reasonable meaning if it is an older iron block Ford or Chev, you should use aluminum heads and other light weight parts. But the handling is just as much about the front to rear distribution of ROLL STIFFNESS as it is about weight distribution. Then there is the effect of front to rear wheel and tire widths, and even tire pressures. A moderate front or rear weight bias can be compensated in several ways, but with the torque of a V8 in one of these cars, you can get oversteer just about any time you want it with a little throttle.
I also think the discussion of engine positioning is pretty academic. If you do a V8 swap, you are going to put the engine about as low and far back as you can, and there is not much room for variation. If you need to butcher the firewall, you probably have chosen the wrong engine.
I believe the constant emphasis on 50-50 weight distribution is a lot of nonsense. The installation of a reasonable V8 will not significantly affect the weight distribution, reasonable meaning if it is an older iron block Ford or Chev, you should use aluminum heads and other light weight parts. But the handling is just as much about the front to rear distribution of ROLL STIFFNESS as it is about weight distribution. Then there is the effect of front to rear wheel and tire widths, and even tire pressures. A moderate front or rear weight bias can be compensated in several ways, but with the torque of a V8 in one of these cars, you can get oversteer just about any time you want it with a little throttle.
I also think the discussion of engine positioning is pretty academic. If you do a V8 swap, you are going to put the engine about as low and far back as you can, and there is not much room for variation. If you need to butcher the firewall, you probably have chosen the wrong engine.
#91
Call me a noob if you will, but even if you do a swap (which i am planning to do very soon), you're not doing it for the entire production line. It's your car. Even if center of gravity and weight distribution is affected, you will compensate for it. A vehicle will adapt to changes you've made to it, for better or worse, and so will you. I can understand wanting to know how customizing will affect a vehicle dynamically, but this is something you have to learn on your own. Each experience is different, what may have worked for someone elses probably won't work for you. Same when you get into high performance, its not easy, so not everyone does it. You guys are pioneers, and need to, for lack of words, keep ******* up until you get it right. A car is nothing without the driver, so I believe its more up to the driver to compensate for a vehicle's lack of perfection.
Last edited by obessius; 06-20-07 at 06:06 PM. Reason: typo
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