ArmyDad
NAXJA Forum User
- Location
- Elk Grove, CA
While doing my home-brew LCoG research, a topic came up periodically. Some folks talked about putting the front axle back into the center of the wheel well after installing a spring lift but not how they arrived at what it took. I thought I might take that on to satisfy my curiosity and use my project as an example. I want to find a way to determine the UCA and LCA measurements that would center the front axle, vertically, in the wheel well.
Now, I made a few assumptions:
1 - The control arms are designed to be parallel to each other (at rest) in the stock condition. It appeared so while I was laying on my back and looking up at my torpedo level while an oil drip from that d**n main seal dropped on my glasses.
2 - I'm only working in the vertical and horizontal planes (no z-axis). There is a z-axis, to some degree, but I chose to ignore it (even us math guys have our limits).
3 - The stock control arms are level (horizontal), the lift will move that arm down (vertical) by the height of the spring. This will give us a right triangle to start with.
4 - People are actually interested in seeing and commenting on intellectual exercises like this.
WARNING --- GEOMETRY ALERT!
==========
A spring lift will rotates the UCA down and back so new UCA would need to be longer than stock LCA. I'm going to be installing a 3" lift so here's what I came up with ...
UCA: Stock length is 15" and I'm installing a 3" spring lift. The new length of the UCA can be calculated by summing the squares of the 2 short sides and taking the square root of the result. In this case ... 15**2 + 3**2 = 234; -/234 = 15.297 or 15.3". In this case, a UCA of 15.3" would put the top of the axle in the same location, vertically, as before the lift.
LCA: Stock length is 15.75" with the same 3" lift. Using the same equation ... (15.75**2) + 3**2 = 257.06; -/257.06 = 16.03". A LCA of 16.0" would put the bottom of the axle in the same location, vertically, as before the lift.
So with my 3" lift an UCA length of 15.3" and a LCA length of 16.0", just over 1/4" each from stock, will put the axle back in the middle of the wheel well vertically.
Keep in mind that now the bottom of the spring would no longer be flexed to the rear but directly beneath the top of the spring. Also, in theory, the castor would change little to none as the axle mounting points for the control arms would have been moved straight down. Reality is obviously different than theory as measurements and threads are not exact to 3 decimal places.
Just looking for some comments so my HS geometry teacher knows I was paying attention.
Now, I made a few assumptions:
1 - The control arms are designed to be parallel to each other (at rest) in the stock condition. It appeared so while I was laying on my back and looking up at my torpedo level while an oil drip from that d**n main seal dropped on my glasses.
2 - I'm only working in the vertical and horizontal planes (no z-axis). There is a z-axis, to some degree, but I chose to ignore it (even us math guys have our limits).
3 - The stock control arms are level (horizontal), the lift will move that arm down (vertical) by the height of the spring. This will give us a right triangle to start with.
4 - People are actually interested in seeing and commenting on intellectual exercises like this.
WARNING --- GEOMETRY ALERT!
==========
A spring lift will rotates the UCA down and back so new UCA would need to be longer than stock LCA. I'm going to be installing a 3" lift so here's what I came up with ...
UCA: Stock length is 15" and I'm installing a 3" spring lift. The new length of the UCA can be calculated by summing the squares of the 2 short sides and taking the square root of the result. In this case ... 15**2 + 3**2 = 234; -/234 = 15.297 or 15.3". In this case, a UCA of 15.3" would put the top of the axle in the same location, vertically, as before the lift.
LCA: Stock length is 15.75" with the same 3" lift. Using the same equation ... (15.75**2) + 3**2 = 257.06; -/257.06 = 16.03". A LCA of 16.0" would put the bottom of the axle in the same location, vertically, as before the lift.
So with my 3" lift an UCA length of 15.3" and a LCA length of 16.0", just over 1/4" each from stock, will put the axle back in the middle of the wheel well vertically.
Keep in mind that now the bottom of the spring would no longer be flexed to the rear but directly beneath the top of the spring. Also, in theory, the castor would change little to none as the axle mounting points for the control arms would have been moved straight down. Reality is obviously different than theory as measurements and threads are not exact to 3 decimal places.
Just looking for some comments so my HS geometry teacher knows I was paying attention.