Would going from 18s to 20s slow me down?
#11
Originally Posted by Veight' post='422671' date='May 9 2007, 03:37 PM
Then why do funny cars have slicks 2 feet wide when they are trying for fast acceleration? Why not use bicycle tires then?
#13
Senior Members
Join Date: Nov 2006
Location: Los Angeles , CA
Posts: 757
Likes: 0
Received 0 Likes
on
0 Posts
slightly slower with the 20's coming from the stock 123 18". I only notice it at the jump. Not a big diff, but Im happy with the looks of the 20's which outweigh the speed.
#15
Originally Posted by jordan23junkee' post='423051' date='May 10 2007, 01:29 PM
slightly slower with the 20's coming from the stock 123 18". I only notice it at the jump. Not a big diff, but Im happy with the looks of the 20's which outweigh the speed.
I will post my dyno sheet I did the other day which really shows how the n52 engine has like NO low-end torque and all the torque is up in the 5-7RPM area-so losing ANY power from a stand is a no no for me. The 18s for me will have to stay! They look great (20s I mean) by the way
#16
Members
Join Date: Dec 2006
Location: UK
Posts: 198
Likes: 0
Received 0 Likes
on
0 Posts
My Ride: Sept '06 535d M sport ,Silver ,Black Dakota Leather,HUD ,TV, Media ,Visibility,Heated F+R seats,Comfort Seats, Active seats ,Sun protection glass,Dark Poplar wood, sunroof ,Elec Sunblinds , Folding mirrors, 19" 172s ,Extended lights Package, Debadged
Originally Posted by rollee' post='423066' date='May 10 2007, 09:22 PM
On another note, will going to a larger size wheel affect the accuracy of the speedo?
Otherwise any change means the speedo needs re-calibrating.
#17
Members
Join Date: Jun 2006
Location: Near Toronto
Posts: 183
Likes: 0
Received 0 Likes
on
0 Posts
My Ride: 06 530Xi w/ M-Sport Package
07 335i Cabriolet w/ M-Sport body kit
Theoretically if the outer diameters of the tires are the same, both assembles weight exactly the same, and the wheels are of the same material and construction differ only in size, the one with a larger wheel inside will accelerate and decelerate slower. This is due to the fact that in the one with a larger wheel inside the weight is more concentrated towards the edge. The higher moment of inertia requires more angular force to accelerate and decelerate.
Moral of the story: If you have to go to larger wheels, buy the very very light ones to offset the negatives of unfavorable weight distribution in the radial direction..
Moral of the story: If you have to go to larger wheels, buy the very very light ones to offset the negatives of unfavorable weight distribution in the radial direction..
#18
Senior Members
Join Date: Nov 2006
Location: Los Angeles , CA
Posts: 757
Likes: 0
Received 0 Likes
on
0 Posts
Originally Posted by sleepyca31' post='423068' date='May 10 2007, 02:23 PM
Ya you obviously do not have a 530 or you would not say a little less off the jump but worth it-trust me as other 530 owners can vouch-we CANNOT afford to lose any off the jump.
I will post my dyno sheet I did the other day which really shows how the n52 engine has like NO low-end torque and all the torque is up in the 5-7RPM area-so losing ANY power from a stand is a no no for me. The 18s for me will have to stay! They look great (20s I mean) by the way
I will post my dyno sheet I did the other day which really shows how the n52 engine has like NO low-end torque and all the torque is up in the 5-7RPM area-so losing ANY power from a stand is a no no for me. The 18s for me will have to stay! They look great (20s I mean) by the way
Sleepy you can go 19" like the 167's or 166's. For performance and speed wise I think thats the max BMW would recommend .
#19
Senior Members
Join Date: Mar 2007
Location: PA, USA
Posts: 351
Likes: 0
Received 0 Likes
on
0 Posts
My Ride: 2004 545i, 1995 M3
Originally Posted by sleepyca31' post='422396' date='May 8 2007, 10:29 PM
So I have a 530 and have 18s on and was looking at 20s and know technically it is like being in a "higher" gear by increasing tire size without compensating in the engine. But is this size diff enough to slow down acceleration. I really cannot afford to lose power in the acceleration as it is piss poor already!
Check out this website to compare sizes.
A taller wheel-and-tire package gives you a higher effective final drive ratio. It's like changing to a higher rear end or transaxle gear (more of a "highway gear," for all you gearheads). This means your vehicle won't accelerate as quickly from a dead stop. On the other hand, the car will achieve a higher top speed (assuming the top speed limiter is removed).
Another thing you need to watch is the weight of your new 20" tire-and-wheel package. Usually most larger diameter wheel-and-tire packages means weigh more then smaller diameter packages. A vehicle's total weight is the sum of all of its parts and affects its ability to accelerate, brake and corner. Reducing the total weight will enhance the vehicle's performance because less weight needs to be controlled and therefore, less energy is required. Unsprung weight is the weight under the springs (wheels, tires, brakes, etc) which moves up and down as the vehicle rides over uneven roads and leans in the corners. Reducing unsprung weight allows the springs and shock absorbers to be more effective in controlling the suspension's movement. Additionally, a vehicle's rotational weight includes all parts that spin including everything in the vehicle's driveline from the engine's crankshaft to its wheels and tires. This affects the energy required to change speed as the vehicle accelerates and brakes. Reducing the weight of any of these rotating components will enhance the vehicle's performance because less energy will be required to increase or decrease their speed. As you would guess, adding weight of any of these rotating components will reduce/degrade the vehicle's performance because more energy will be required to increase or decrease their speed. It is commonly known that each pound of rotational mass lost (unsprung weight) provides an equivalent performance gain as a 10 pound reduction in vehicle weight. Thus if are new 20" package weighs 15 pounds more on each corner (15 X 4 X 10 = 600) it will be like adding 600 pounds to your car, not good for performance.
#20
Originally Posted by Bad Bimr' post='423642' date='May 12 2007, 06:50 AM
If sized correctly your 20" wheel-and-tire package can be the same size (circumfrence, diameter, and revolutions per miles) as you 18" package.
Check out this website to compare sizes.
A taller wheel-and-tire package gives you a higher effective final drive ratio. It's like changing to a higher rear end or transaxle gear (more of a "highway gear," for all you gearheads). This means your vehicle won't accelerate as quickly from a dead stop. On the other hand, the car will achieve a higher top speed (assuming the top speed limiter is removed).
Another thing you need to watch is the weight of your new 20" tire-and-wheel package. Usually most larger diameter wheel-and-tire packages means weigh more then smaller diameter packages. A vehicle's total weight is the sum of all of its parts and affects its ability to accelerate, brake and corner. Reducing the total weight will enhance the vehicle's performance because less weight needs to be controlled and therefore, less energy is required. Unsprung weight is the weight under the springs (wheels, tires, brakes, etc) which moves up and down as the vehicle rides over uneven roads and leans in the corners. Reducing unsprung weight allows the springs and shock absorbers to be more effective in controlling the suspension's movement. Additionally, a vehicle's rotational weight includes all parts that spin including everything in the vehicle's driveline from the engine's crankshaft to its wheels and tires. This affects the energy required to change speed as the vehicle accelerates and brakes. Reducing the weight of any of these rotating components will enhance the vehicle's performance because less energy will be required to increase or decrease their speed. As you would guess, adding weight of any of these rotating components will reduce/degrade the vehicle's performance because more energy will be required to increase or decrease their speed. It is commonly known that each pound of rotational mass lost (unsprung weight) provides an equivalent performance gain as a 10 pound reduction in vehicle weight. Thus if are new 20" package weighs 15 pounds more on each corner (15 X 4 X 10 = 600) it will be like adding 600 pounds to your car, not good for performance.
Check out this website to compare sizes.
A taller wheel-and-tire package gives you a higher effective final drive ratio. It's like changing to a higher rear end or transaxle gear (more of a "highway gear," for all you gearheads). This means your vehicle won't accelerate as quickly from a dead stop. On the other hand, the car will achieve a higher top speed (assuming the top speed limiter is removed).
Another thing you need to watch is the weight of your new 20" tire-and-wheel package. Usually most larger diameter wheel-and-tire packages means weigh more then smaller diameter packages. A vehicle's total weight is the sum of all of its parts and affects its ability to accelerate, brake and corner. Reducing the total weight will enhance the vehicle's performance because less weight needs to be controlled and therefore, less energy is required. Unsprung weight is the weight under the springs (wheels, tires, brakes, etc) which moves up and down as the vehicle rides over uneven roads and leans in the corners. Reducing unsprung weight allows the springs and shock absorbers to be more effective in controlling the suspension's movement. Additionally, a vehicle's rotational weight includes all parts that spin including everything in the vehicle's driveline from the engine's crankshaft to its wheels and tires. This affects the energy required to change speed as the vehicle accelerates and brakes. Reducing the weight of any of these rotating components will enhance the vehicle's performance because less energy will be required to increase or decrease their speed. As you would guess, adding weight of any of these rotating components will reduce/degrade the vehicle's performance because more energy will be required to increase or decrease their speed. It is commonly known that each pound of rotational mass lost (unsprung weight) provides an equivalent performance gain as a 10 pound reduction in vehicle weight. Thus if are new 20" package weighs 15 pounds more on each corner (15 X 4 X 10 = 600) it will be like adding 600 pounds to your car, not good for performance.
This may be way off base but does the same log apply if you REDUCE weight on each corner by 15 pounds-is that like a "600" pound weight loss in terms of acceleration?-if so that would be my next mod considering the Z4 that weighs 400 less seemingly accelerates 1 second faster in the 0-60-sign me up! Unsprung weight-a great find!