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sounds silly, but...hard to raise M3 using scissor jack?

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I'm unclear how suspension plays into this but you are certainly right that until the tyres come off the ground some of the car's weight rests on them.
If there was no suspension then the you'd be lifting the entire weight of the car right away. The softer the suspension the less force it takes when you first start lifting the car. Imagine trying to lift a 4000lb block of steel, no way it would even budge. However if you try to lift a car you can lift it an inch or two.
 
If there was no suspension then the you'd be lifting the entire weight of the car right away. The softer the suspension the less force it takes when you first start lifting the car. Imagine trying to lift a 4000lb block of steel, no way it would even budge. However if you try to lift a car you can lift it an inch or two.
  • One corner only
  • I'm not sure about the leverage but an 8 inch handle on a scissor jack would travel about 25 inches in a rotation and ?? perhaps raise the car an inch
 
How do you use a jackpad adapter with a scissor jack?

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Huh. I'll have to pull mine out tomorrow. I keep seeing the opposite in my head

Addendum:
The jack forms a right triangle, and the hypotenuse does not change.
As we rotate the arm and the jack rises the x value shortens and the y value increases but the relation
x^2 + y^2 = R^2 stays constant. The x shortening per rotation is constant.

Say we start at x = 6 and y = 1, then R = sqrt(37)
If we shorten x by 1 then x = 5 and y = sqrt (37 - 25) = sqrt (12)
If we shorten x another 1 then x = 4 and y = sqrt (37 - 16) = sqrt (21)
And another 1 for good measure: now x = 3 and y = sqrt (37 -9) = sqrt (28)

The vertical gain per decreases increases as the angle opens.

You were close on your math but you just lost scope of the problem.
You need to look at the delta Y for each unit change in X.
You‘re calculating the total Y on each iteration.

Addendum 2:
Let’s do the last 2 steps (jack fully up)

X = 1 and y = SQRT (37 - 1) = 6
X = 0 and y = SQRT (37 - 0) = 6.08 (delta Y is a tiny 0.08 for delta X of 1)

Let’s go back to your first step

Y = 1 (starting Y, jack fully down)
Y = SQRT (12) = 3.46 (delta Y is a huge 2.24 for delta X of 1)
 
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Which scissor jack do you have? Looking for one to keep in the car along with my plug kit.

I'm curious as well. I'm looking for a scissor jack to keep in the car along with tire repair kit. Also, would like to know if I buy the scissor jack, I wouldn't need a separate wrench to unscrew the wheels? I was initially thinking about going for a full spare kit from modernspare.com but what are the actual chances of using the spare? I never had to use it in past 20+ years..

So, I just need some advice on which torque wrench to unscrew the wheels off and a strong scissor jack. Thanks in advance!
 
I'm curious as well. I'm looking for a scissor jack to keep in the car along with tire repair kit. Also, would like to know if I buy the scissor jack, I wouldn't need a separate wrench to unscrew the wheels? I was initially thinking about going for a full spare kit from modernspare.com but what are the actual chances of using the spare? I never had to use it in past 20+ years..

So, I just need some advice on which torque wrench to unscrew the wheels off and a strong scissor jack. Thanks in advance!

First post describes the jack I use, scroll down a few posts and links for wrench, socket etc.

The perfect Model 3 emergency jack?