There is a woman I really like. I was with her the other night. She is 5'8" and 300 lbs. I am 6'1" and 200 lbs (I am not fat - muscular). She was fantastic in bed. When we did it missionary and doggie style the sex was great. She wanted to be on top but I was a little scared the weight would be too much. I know her full weight won't be on me but I am still unsure. What do you think?
2007-08-09 03:57:56 · 13 answers · asked by Handsome Stranger 2 in Family & Relationships ➔ Singles & Dating
I dont think it would be a problem. Just let her try it out and see how it goes... and if its not working then just tell her you like the other way.
2007-08-09 04:12:43 · answer #1 · answered by Chelley 3 · 0⤊ 0⤋
Depends -My ex was around the 270's. I'm 125. I was always on top and doing all the work. Hated it. In this case, it did matter!
2007-08-09 11:04:27 · answer #2 · answered by Anonymous · 0⤊ 0⤋
For me, it has to. I am 5'0, 105 pounds, and I cannot have some 250 lbs. guy on top of me. I would suffocate!
2007-08-09 11:01:36 · answer #3 · answered by This Is Radio Clash 6 · 0⤊ 0⤋
I'd stick to some thing that isn't going to kill you like doggie style
2007-08-09 11:07:13 · answer #4 · answered by flkenout 3 · 1⤊ 0⤋
My man, BBWs are so grateful to get a muscle man in bed they do go for it. As for her weight, you're safe with her on top.
2007-08-09 11:02:48 · answer #5 · answered by Net Rider 3 · 0⤊ 0⤋
Whatever works for you hun. I say go for it. Be a man. Take the weight.
2007-08-09 11:02:01 · answer #6 · answered by Charlette 2 · 0⤊ 0⤋
if the sex is great, there wouldn't be a problem try it you never know..
2007-08-09 11:02:42 · answer #7 · answered by comfy c 6 · 0⤊ 0⤋
you wont be weighed down by her because most of her weight wont be on you. you'll love it. my guy does and i weigh more than your girl.
2007-08-09 11:02:20 · answer #8 · answered by Victoria J 3 · 0⤊ 0⤋
LOL
be careful...dont let her squish u
2007-08-09 11:04:18 · answer #9 · answered by jane doe 3 · 1⤊ 0⤋
hahaa
2007-08-09 12:04:34 · answer #10 · answered by Anonymous · 0⤊ 0⤋