10 plate hack squat for reps! http://www.youtube.com/watch?v=6i1izPR3pC8

[BigNick16]

thnx. link in title or you can go to my youtube page =)

hope you like

[FunkymonkAW]

[PeterGibbons316]

[Chicago712]

machine? come on bra

step yo game up

[IDrinkBloodLOL]

A hack squat is a deadlift done behind the back, not a machine.



Secondly, a "10 plate squat" of any kind means 10 plates on each side. You have a 5 plate machine hack squat.

Lastly, since the machine is at a 45 degree angle, the maximum amount of force you could be pushing is ~71% of the weighted of the plates. That's from physics*. So, what you really get is 10 x 45 pounds = 450 pounds x 0.71 = about 320.

Congratulations OP, you squat 3 plates!

*Math is as follows:
F is the weight of all plates combined. Its vector points straight down. If we use a normal vertical coordinate system, Fx is zero and Fy=F.

The force is broken into component vectors, Fparallel and Fnormal, which run parallel and perpendicular to the path of the sled on the machine. The component parallel to the sled is the force you actually push against. These machines are at 45 degrees, so Fnormal=Fparallel and Fy=(Fnormal + Fparallel)/(sqrt 2)=(2 Fnormal)/(sqrt 2)=(sqrt 2)(Fnormal). Therefore, Fy/(sqrt 2)=Fnormal. Since Fy is the weight of the plates, and 1/(sqrt 2) is about 0.71, you only need to push 71% as hard as the total weight loaded to move it. Thus, 71% of 450 = 320.

gg.

[Chicago712]

A hack squat is a deadlift done behind the back, not a machine.



Secondly, a "10 plate squat" of any kind means 10 plates on each side. You have a 5 plate machine hack squat.

Lastly, since the machine is at a 45 degree angle, the maximum amount of force you could be pushing is ~71% of the weighted of the plates. That's from physics. So, what you really get is 10 x 45 pounds = 450 pounds x 0.71 = about 320.

Congratulations OP, you squat 3 plates!

thank you rep+

[qeynos]

i barbell hack squat 330 for 12 reps 20 times a second for 50 years

[BombDonald]

A hack squat is a deadlift done behind the back, not a machine.



Secondly, a "10 plate squat" of any kind means 10 plates on each side. You have a 5 plate machine hack squat.

Lastly, since the machine is at a 45 degree angle, the maximum amount of force you could be pushing is ~71% of the weighted of the plates. That's from physics*. So, what you really get is 10 x 45 pounds = 450 pounds x 0.71 = about 320.

Congratulations OP, you squat 3 plates!

*Math is as follows:
F is the weight of all plates combined. Its vector points straight down. If we use a normal vertical coordinate system, Fx is zero and Fy=F.

The force is broken into component vectors, Fparallel and Fnormal, which run parallel and perpendicular to the path of the sled on the machine. The component parallel to the sled is the force you actually push against. These machines are at 45 degrees, so Fnormal=Fparallel and Fy=(Fnormal + Fparallel)/(sqrt 2)=(2 Fnormal)/(sqrt 2)=(sqrt 2)(Fnormal). Therefore, Fy/(sqrt 2)=Fnormal. Since Fy is the weight of the plates, and 1/(sqrt 2) is about 0.71, you only need to push 71% as hard as the total weight loaded to move it. Thus, 71% of 450 = 320.

gg.

[chazzy1864]

A hack squat is a deadlift done behind the back, not a machine.



Secondly, a "10 plate squat" of any kind means 10 plates on each side. You have a 5 plate machine hack squat.

Lastly, since the machine is at a 45 degree angle, the maximum amount of force you could be pushing is ~71% of the weighted of the plates. That's from physics*. So, what you really get is 10 x 45 pounds = 450 pounds x 0.71 = about 320.

Congratulations OP, you squat 3 plates!

*Math is as follows:
F is the weight of all plates combined. Its vector points straight down. If we use a normal vertical coordinate system, Fx is zero and Fy=F.

The force is broken into component vectors, Fparallel and Fnormal, which run parallel and perpendicular to the path of the sled on the machine. The component parallel to the sled is the force you actually push against. These machines are at 45 degrees, so Fnormal=Fparallel and Fy=(Fnormal + Fparallel)/(sqrt 2)=(2 Fnormal)/(sqrt 2)=(sqrt 2)(Fnormal). Therefore, Fy/(sqrt 2)=Fnormal. Since Fy is the weight of the plates, and 1/(sqrt 2) is about 0.71, you only need to push 71% as hard as the total weight loaded to move it. Thus, 71% of 450 = 320.

gg.

lol damn

[Xuaxace]

A hack squat is a deadlift done behind the back, not a machine.



Secondly, a "10 plate squat" of any kind means 10 plates on each side. You have a 5 plate machine hack squat.

Lastly, since the machine is at a 45 degree angle, the maximum amount of force you could be pushing is ~71% of the weighted of the plates. That's from physics*. So, what you really get is 10 x 45 pounds = 450 pounds x 0.71 = about 320.

Congratulations OP, you squat 3 plates!

*Math is as follows:
F is the weight of all plates combined. Its vector points straight down. If we use a normal vertical coordinate system, Fx is zero and Fy=F.

The force is broken into component vectors, Fparallel and Fnormal, which run parallel and perpendicular to the path of the sled on the machine. The component parallel to the sled is the force you actually push against. These machines are at 45 degrees, so Fnormal=Fparallel and Fy=(Fnormal + Fparallel)/(sqrt 2)=(2 Fnormal)/(sqrt 2)=(sqrt 2)(Fnormal). Therefore, Fy/(sqrt 2)=Fnormal. Since Fy is the weight of the plates, and 1/(sqrt 2) is about 0.71, you only need to push 71% as hard as the total weight loaded to move it. Thus, 71% of 450 = 320.

gg.

<3.

[netopia]

A hack squat is a deadlift done behind the back, not a machine.



Secondly, a "10 plate squat" of any kind means 10 plates on each side. You have a 5 plate machine hack squat.

Lastly, since the machine is at a 45 degree angle, the maximum amount of force you could be pushing is ~71% of the weighted of the plates. That's from physics*. So, what you really get is 10 x 45 pounds = 450 pounds x 0.71 = about 320.

Congratulations OP, you squat 3 plates!

*Math is as follows:
F is the weight of all plates combined. Its vector points straight down. If we use a normal vertical coordinate system, Fx is zero and Fy=F.

The force is broken into component vectors, Fparallel and Fnormal, which run parallel and perpendicular to the path of the sled on the machine. The component parallel to the sled is the force you actually push against. These machines are at 45 degrees, so Fnormal=Fparallel and Fy=(Fnormal + Fparallel)/(sqrt 2)=(2 Fnormal)/(sqrt 2)=(sqrt 2)(Fnormal). Therefore, Fy/(sqrt 2)=Fnormal. Since Fy is the weight of the plates, and 1/(sqrt 2) is about 0.71, you only need to push 71% as hard as the total weight loaded to move it. Thus, 71% of 450 = 320.

gg.

Brilliance, sheer brilliance

[ZoranM]

I bet all of you can hack machine squat 450 for reps.

[Engineer_Guy]

OP didn't even provide a good camera angle of his accomplishment!

[BombDonald]

I remember this guy now after looking through some YouTube videos.

He got torched in the powerlifting section a while back for a video of a 500x4 HIGH squat that looked like he was teabagging the camera.

http://forum.bodybuilding.com/showth...hp?t=132666263

[im2manly]

A hack squat is a deadlift done behind the back, not a machine.



Secondly, a "10 plate squat" of any kind means 10 plates on each side. You have a 5 plate machine hack squat.

Lastly, since the machine is at a 45 degree angle, the maximum amount of force you could be pushing is ~71% of the weighted of the plates. That's from physics*. So, what you really get is 10 x 45 pounds = 450 pounds x 0.71 = about 320.

Congratulations OP, you squat 3 plates!

*Math is as follows:
F is the weight of all plates combined. Its vector points straight down. If we use a normal vertical coordinate system, Fx is zero and Fy=F.

The force is broken into component vectors, Fparallel and Fnormal, which run parallel and perpendicular to the path of the sled on the machine. The component parallel to the sled is the force you actually push against. These machines are at 45 degrees, so Fnormal=Fparallel and Fy=(Fnormal + Fparallel)/(sqrt 2)=(2 Fnormal)/(sqrt 2)=(sqrt 2)(Fnormal). Therefore, Fy/(sqrt 2)=Fnormal. Since Fy is the weight of the plates, and 1/(sqrt 2) is about 0.71, you only need to push 71% as hard as the total weight loaded to move it. Thus, 71% of 450 = 320.

gg.

damn will rep once off RC

[chazzy1864]

I remember this guy now after looking through some YouTube videos.

He got torched in the powerlifting section a while back for a video of a 500x4 HIGH squat that looked like he was teabagging the camera.

http://forum.bodybuilding.com/showth...hp?t=132666263

Wow, I'm still not through that thread and so far, pretty every post of his I want to neg. I'm debating since it was from a year ago.

[VoxExMachina]

I wonder how many people can machine hack squat as much as they can barbell squat.

[GriPz]

lol thats 5 plaates... I squat 192.5kg for 8 reps. Be a man and dont use machine

[ZoranM]

Be a man and dont use machine

I'm more than happy to provide you with a gift that has the color of a womans period.

[ironwill2008]

LOL @ "10 plates."

I remember this guy now after looking through some YouTube videos.

He got torched in the powerlifting section a while back for a video of a 500x4 HIGH squat that looked like he was teabagging the camera.

http://forum.bodybuilding.com/showth...hp?t=132666263

Lulz. That thread is hilarious!

Repz when off spread.

I wonder how many people can machine hack squat as much as they can barbell squat.

I know I can, and I'm 1000 years old.

[pezking7p]

*Math is as follows:[size=-2]
F is the weight of all plates combined. Its vector points straight down. If we use a normal vertical coordinate system, Fx is zero and Fy=F.

The force is broken into component vectors, Fparallel and Fnormal, which run parallel and perpendicular to the path of the sled on the machine. The component parallel to the sled is the force you actually push against. These machines are at 45 degrees, so Fnormal=Fparallel and Fy=(Fnormal + Fparallel)/(sqrt 2)=(2 Fnormal)/(sqrt 2)=(sqrt 2)(Fnormal). Therefore, Fy/(sqrt 2)=Fnormal. Since Fy is the weight of the plates, and 1/(sqrt 2) is about 0.71, you only need to push 71% as hard as the total weight loaded to move it. Thus, 71% of 450 = 320.

gg.[size]

Bro, next time just use 450*cos(45)=320 and save us some squinting. Also, you forgot frictional forces and the weight of the sled.

[creationd]

machine? come on bra

step yo game up

Have you ever used the hack squat? The only reason it seems easy to most people because like on the leg press most people do not go full range of motion, I struggled to hack squat as much as I could barbell squat. But in this vid it seems his ass is not really going down below his knees. But hard to tell from this vid.

[BigNick16]

Congratulations OP, you squat 3 plates!

[/size][/i]

yea, i actually squat 3 plates, each side, for 24 reps

[BigNick16]

Have you ever used the hack squat? The only reason it seems easy to most people because like on the leg press most people do not go full range of motion, I struggled to hack squat as much as I could barbell squat. But in this vid it seems his ass is not really going down below his knees. But hard to tell from this vid.

you must have horrible eyes, because it is clearly going FAR below my knees

[creationd]

you must have horrible eyes, because it is clearly going FAR below my knees

Clearly? What vid are you watching in your vid the knees and butt go out of frame at each rep.

[Destructodave]

Clearly? What vid are you watching in your vid the knees and butt go out of frame at each rep.

Honestly guy, you can see the sled bottoming out on every rep but the first 1. It looks as if he is going as low as the sled allows, and even using it to bounce the weight off. So, I'm pretty sure hes going to at least knee level.

[StrappingYngLad]

This man is no man, he is a beast. Groowwwll!

[vollric]

A hack squat is a deadlift done behind the back, not a machine.



Secondly, a "10 plate squat" of any kind means 10 plates on each side. You have a 5 plate machine hack squat.

Lastly, since the machine is at a 45 degree angle, the maximum amount of force you could be pushing is ~71% of the weighted of the plates. That's from physics*. So, what you really get is 10 x 45 pounds = 450 pounds x 0.71 = about 320.

Congratulations OP, you squat 3 plates!

*Math is as follows:
F is the weight of all plates combined. Its vector points straight down. If we use a normal vertical coordinate system, Fx is zero and Fy=F.

The force is broken into component vectors, Fparallel and Fnormal, which run parallel and perpendicular to the path of the sled on the machine. The component parallel to the sled is the force you actually push against. These machines are at 45 degrees, so Fnormal=Fparallel and Fy=(Fnormal + Fparallel)/(sqrt 2)=(2 Fnormal)/(sqrt 2)=(sqrt 2)(Fnormal). Therefore, Fy/(sqrt 2)=Fnormal. Since Fy is the weight of the plates, and 1/(sqrt 2) is about 0.71, you only need to push 71% as hard as the total weight loaded to move it. Thus, 71% of 450 = 320.

gg.

pwned!

[gkicksass]

im probably gonna get flamed for this but a 5 plate machine hack squat is decent lift IMO. however if you look at the video again it kinda looks like hes bouncing off at the bottom. maybe its just me but it looks kinda weird.

btw op, negged cus youre an arrogant little sht from your other thread bombdonald posted.

[Engineer_Guy]

im probably gonna get flamed for this but a 5 plate machine hack squat is decent lift IMO.

Depends on your definition of decent...also remember he's on an angled sled so the actual amount he's lifting is quite a bit less than 5 plates.