A Furious 7 physics question [spoilers]
April 13, 2015 8:42 AM Subscribe
After seeing Furious 7 this weekend, a physics question came up about weight and leverage.
Spoilers and such below the break.
There is a scene where Vin Diesel's character deadlifts the front end of a Lykan Hypersport and holds it for a few minutes while Paul Walker pokes around underneath. The implication is that 1) Vin Diesel is super strong and 2) this supercar is super light.
The Hypersport has a curb weight of 1380kg (3042lbs), and the movie version was "bullet-proof" so it was probably even heavier. The world record for a deadlift is 1180lbs. Considering that the car is rear engine with the weight distributed more to the back, and the he was only lifting the front of the car with the fulcrum at the rear axle, how much weight was Vin supposedly lifting in that scene?
Some dimensions for the Hypersport:
Length: 4480 mm (175 in)
Width: 1944 mm (76.5 in)
Height: 1170 mm (45 in)
Wheelbase: 2625 mm (103.3 in)
Front Track: 1506 mm (59.3 in)
Back Track: 1585 mm (62.4 in)
Estimated Kerb Weight: 1380 kg
There is a scene where Vin Diesel's character deadlifts the front end of a Lykan Hypersport and holds it for a few minutes while Paul Walker pokes around underneath. The implication is that 1) Vin Diesel is super strong and 2) this supercar is super light.
The Hypersport has a curb weight of 1380kg (3042lbs), and the movie version was "bullet-proof" so it was probably even heavier. The world record for a deadlift is 1180lbs. Considering that the car is rear engine with the weight distributed more to the back, and the he was only lifting the front of the car with the fulcrum at the rear axle, how much weight was Vin supposedly lifting in that scene?
Some dimensions for the Hypersport:
Length: 4480 mm (175 in)
Width: 1944 mm (76.5 in)
Height: 1170 mm (45 in)
Wheelbase: 2625 mm (103.3 in)
Front Track: 1506 mm (59.3 in)
Back Track: 1585 mm (62.4 in)
Estimated Kerb Weight: 1380 kg
Without knowing the weight distribution, it's only a guess.
That said, my brother in law picked up the back of a Mustang that had me pinned. Using specs from here, and assuming the rear overhang is 24" more than the front over hang, he lifted 1400 pounds. Now, that's not a from-the-ground deadlift but rather a six-inch lift from an almost-standing position.
He's a fucking hoss tho.
posted by notsnot at 9:12 AM on April 13, 2015 [2 favorites]
That said, my brother in law picked up the back of a Mustang that had me pinned. Using specs from here, and assuming the rear overhang is 24" more than the front over hang, he lifted 1400 pounds. Now, that's not a from-the-ground deadlift but rather a six-inch lift from an almost-standing position.
He's a fucking hoss tho.
posted by notsnot at 9:12 AM on April 13, 2015 [2 favorites]
Related: Someone else has been thinking about the physics in this movie: Could Furious 7’s Double-Skyscraper Jump Really Happen? We Asked a Physicist
posted by bassomatic at 9:38 AM on April 13, 2015 [2 favorites]
posted by bassomatic at 9:38 AM on April 13, 2015 [2 favorites]
A back of an envelope calculation to approximately work this out isn't too difficult. You need to work out the moment (turning force) downwards, and as the car is in mechanical equilibrium (i.e it isn't moving), this will be equal to the upwards moment, form which we can get the force.
A moment is equal to the force multiplied by the length from the pivot, so the downwards moment = L1 * Fg (length * weight) = L1 * m * g (length to centre of mass of car * mass * gravitational constant)
This is equal to the upwards moment = L2 * Fvd (length * force Vin is exerting)
Setting them equal to each other and rearranging, Fvd = (L1 * m * g)/L2
Using the value for mass you give, g = 10, and guessing Vin Diesel is twice as far from the pivot as the centre of mass of the car is (L2 = 2*L1) we find a force of 6900 N, which is like lifting 700kg. Playing around with the distances will give you slightly different answers.
N.B having written all of that out, I can see it's only the ratio of the distances L1 and L2 that matters (which I should have remembered straight away) - here as L2 = 2*L1 the weight Vin has to be able to lift is half the weight of the car, so that's the quick way of calculating it.
posted by Ned G at 9:59 AM on April 13, 2015 [1 favorite]
A moment is equal to the force multiplied by the length from the pivot, so the downwards moment = L1 * Fg (length * weight) = L1 * m * g (length to centre of mass of car * mass * gravitational constant)
This is equal to the upwards moment = L2 * Fvd (length * force Vin is exerting)
Setting them equal to each other and rearranging, Fvd = (L1 * m * g)/L2
Using the value for mass you give, g = 10, and guessing Vin Diesel is twice as far from the pivot as the centre of mass of the car is (L2 = 2*L1) we find a force of 6900 N, which is like lifting 700kg. Playing around with the distances will give you slightly different answers.
N.B having written all of that out, I can see it's only the ratio of the distances L1 and L2 that matters (which I should have remembered straight away) - here as L2 = 2*L1 the weight Vin has to be able to lift is half the weight of the car, so that's the quick way of calculating it.
posted by Ned G at 9:59 AM on April 13, 2015 [1 favorite]
Note that 'bullet-proofing' will add anything from 400 lb to over 2,000 lb to the weight of the car (the higher figure would handle armour-piercing rounds).
posted by pipeski at 10:28 AM on April 13, 2015
posted by pipeski at 10:28 AM on April 13, 2015
The hardest part of the deadlift is getting it off the ground, though. You'd be able to lift more with a smaller range of motion than a normal deadlift, which starts 8 inches off the ground and ends well above the knee.
posted by wnissen at 9:24 PM on April 13, 2015
posted by wnissen at 9:24 PM on April 13, 2015
To add some real world stuff here - the weight distribution of that car is at best (in the favour of Master Diesel) 60% rear, 40% front with no-one in it. More likely 56/44 or 58/42 or it'd handle like poop. Although the most important thing is that the vast majority of the weight is forward of the rear axle (probably only part of the transmission and some light bodywork behind it) so very little of the weight of the car is helping him (ie acting downwards of the rear axle to help his upwards lifting of the front axle). So he's lifting the majority of the weight of the car with the C of G maybe 1 metre or so in front of the rear axle.
I'd concur that you're looking at easily 4-600lbs for bullet proofing of the car, which ups the weight.
So he is lifting a LOT of weight (maybe 3,500lbs) with the leverage from Ned G's answer meaning that he is maybe 2m from that C of G so it is a lot less than the overall weight of the car but still *Cough*bullshit*cough* high levels of load.
Also; there is nothing at all strong enough at the front of any car - especially one like that - that would allow him to lift a fraction of that weight. Assuming his strength wasn't an issue (arf) then the most that'd happen is he'd stand up with some broken bits of bumper/grill/intake and bodywork in his hand. Cars don't have handles at the front and you'd need to be maybe a foot in from the front of the car before there is any substantial part of the chassis you could use to lift the whole front end of the car with.
posted by Brockles at 6:19 AM on April 14, 2015 [1 favorite]
I'd concur that you're looking at easily 4-600lbs for bullet proofing of the car, which ups the weight.
So he is lifting a LOT of weight (maybe 3,500lbs) with the leverage from Ned G's answer meaning that he is maybe 2m from that C of G so it is a lot less than the overall weight of the car but still *Cough*bullshit*cough* high levels of load.
Also; there is nothing at all strong enough at the front of any car - especially one like that - that would allow him to lift a fraction of that weight. Assuming his strength wasn't an issue (arf) then the most that'd happen is he'd stand up with some broken bits of bumper/grill/intake and bodywork in his hand. Cars don't have handles at the front and you'd need to be maybe a foot in from the front of the car before there is any substantial part of the chassis you could use to lift the whole front end of the car with.
posted by Brockles at 6:19 AM on April 14, 2015 [1 favorite]
Assuming the following:
1. The car's center of mass is located halfway between the wheelbase.
2. The midpoint of the wheelbase and midpoint of the car's length are coincident.
3. Vin is lifting from the front bumber of the car and the rear wheel stays in contact with the ground.
The lifting force to initially get the front wheel off of the ground is given by a moment balance about the rear wheel:
Moment at rear wheel = Mrw
Length of wheelbase = Lwb
Length of car = Lcr
Mass of car = m
Gravitational constant = g = 9.8 m/s^2
Lift force = F
Mrw = 0 = (Lwb/2)*(-m*g)+(Lcr/2+Lwb/2)*F
rearranging and solving for F... F=(Lwb/(Lcr+Lwb))*m*g
plugging in the values F=(2.625/(4.480+2.625))*1380*9.8 = 5000N or 1124lbf
The big assumption made in these calculations is that the center of mass of the car is at the geometric center of the car. In reality, I would expect that the center of mass would be skewed towards the front of the car where the engine block resides. This would reduce the mechanical advantage and result in a higher force than computed here.
posted by incolorinred at 1:20 PM on April 14, 2015
1. The car's center of mass is located halfway between the wheelbase.
2. The midpoint of the wheelbase and midpoint of the car's length are coincident.
3. Vin is lifting from the front bumber of the car and the rear wheel stays in contact with the ground.
The lifting force to initially get the front wheel off of the ground is given by a moment balance about the rear wheel:
Moment at rear wheel = Mrw
Length of wheelbase = Lwb
Length of car = Lcr
Mass of car = m
Gravitational constant = g = 9.8 m/s^2
Lift force = F
Mrw = 0 = (Lwb/2)*(-m*g)+(Lcr/2+Lwb/2)*F
rearranging and solving for F... F=(Lwb/(Lcr+Lwb))*m*g
plugging in the values F=(2.625/(4.480+2.625))*1380*9.8 = 5000N or 1124lbf
The big assumption made in these calculations is that the center of mass of the car is at the geometric center of the car. In reality, I would expect that the center of mass would be skewed towards the front of the car where the engine block resides. This would reduce the mechanical advantage and result in a higher force than computed here.
posted by incolorinred at 1:20 PM on April 14, 2015
I would expect that the center of mass would be skewed towards the front of the car where the engine block resides.
It's a mid engined car (behind the driver). That configuration usually results in the kind of weight distribution that I mentioned above.
posted by Brockles at 2:07 PM on April 14, 2015
It's a mid engined car (behind the driver). That configuration usually results in the kind of weight distribution that I mentioned above.
posted by Brockles at 2:07 PM on April 14, 2015
Best answer: Revising for the 60/40 mass distribution (moves the CM by 10% of the total car length towards back wheel) and adding bullet proofing of 600lbm or 272kg to the car's mass; the force is about the same (slightly lower):
F=((2.625-0.4480)/(4.480+2.625))*(1380+272)*9.8
F=4,960N or 1,115lbf
Doable for a human being, albiet a super ripped one.
posted by incolorinred at 9:11 PM on April 14, 2015 [1 favorite]
F=((2.625-0.4480)/(4.480+2.625))*(1380+272)*9.8
F=4,960N or 1,115lbf
Doable for a human being, albiet a super ripped one.
posted by incolorinred at 9:11 PM on April 14, 2015 [1 favorite]
Response by poster: So it's not impossible, it would just take Vin's character being one of the strongest people on the planet and that car having a specially reinforced front bumper that can hold the weight. Works for me!
posted by thecjm at 12:27 PM on April 15, 2015
posted by thecjm at 12:27 PM on April 15, 2015
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Ok, assuming the weight is concentrated at a point 1/4 the distance of the total length, he would be lifting about 345kg or 759lbs.
posted by spikeleemajortomdickandharryconnickjrmints at 9:04 AM on April 13, 2015 [2 favorites]